[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix

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mb5fcdf3c3c009f 2021-08-18 21:16:00

文章标签 Medium 二维 ide i++ 二分查找 文章分类 代码人生

leetcode74

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_Medium_02

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

二维矩阵里找目标值target

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

【思路】

二分

每行单独二分查找，找到直接返回true，没找到最后返回false

【代码】

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int i=0;i<matrix.length;i++){
            int left=0;int right=matrix[0].length;
            while(left<right){
                int mid=left+(right-left)/2;
                if(matrix[i][mid]==target){
                    return true;
                }
                else if(matrix[i][mid]<target){
                    left=mid+1;
                }
                else if(matrix[i][mid]>target){
                    right=mid;
                }
                    
            }
        }
        return false;
    }
}

leetcode240

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}

【题目】

变式，同样是二维矩阵里找目标值target，但满足上到下，左到右均递增

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二分查找_03

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_04

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

【思路】

可以直接用74的方法解，但时间较久

这里考虑到上下、左右递增，从左下/右上角出发遍历即可

每个点都有两种大/小了的选择

左上同大，右下同小，不行

B站https://www.bilibili.com/video/BV1Dy4y1u7RH 讲的很清楚

[Leetcode 74/240]二维数组中找目标值Search a 2D Matrix_二维_05

【代码】

public static boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length < 1 || matrix[0].length < 1)
        return false;

    int j = 0;
    int i= matrix.length - 1;
    while (j <= matrix[0].length - 1 && i>= 0) {
        if (target == matrix[i][j])
            return true;
        else if (target < matrix[i][j])
            i--;
        else if (target > matrix[i][j])
            j++;
    }
    return false;
}