目录

​​1，题目描述​​

​​2，思路​​

​​3，代码【C++】​​

1，题目描述

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, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false


链接：https://leetcode-cn.com/problems/search-a-2d-matrix
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2，思路

把整个矩阵看成一个一维数组，对这个一维数组进行二分查找；

 

3，代码【C++】

官方题解：

class Solution {
  public:
  bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int m = matrix.size();
    if (m == 0) return false;
    int n = matrix[0].size();

    // 二分查找
    int left = 0, right = m * n - 1;
    int pivotIdx, pivotElement;
    while (left <= right) {
      pivotIdx = (left + right) / 2;
      pivotElement = matrix[pivotIdx / n][pivotIdx % n];
      if (target == pivotElement) return true;
      else {
        if (target < pivotElement) right = pivotIdx - 1;
        else left = pivotIdx + 1;
      }
    }
    return false;
  }
};