[LeetCode] 54. Spiral Matrix

Given an m x n matrix, return all elements of the matrix in spiral order.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

螺旋矩阵。

题意是给一个二维数组，请你螺旋遍历input，输出一个一维数组记录螺旋遍历的结果。

这个题我没有什么比较巧妙的思路，只能照着遍历的方式去实现了代码，按照右 - 下 - 左 - 上的顺序去遍历。这个题做的时候需要用几个变量去记录二维数组坐标的边界，同时记得当每一行/列遍历完了之后，边界值需要减1因为遍历的时候每一圈的size都在减小。

时间O(mn)

空间O(n) - 输出是一个一维数组

JavaScript实现

 1 /**
 2  * @param {number[][]} matrix
 3  * @return {number[]}
 4  */
 5 var spiralOrder = function (matrix) {
 6     let res = [];
 7     if (matrix.length === 0) {
 8         return res;
 9     }
10     let rowBegin = 0;
11     let rowEnd = matrix.length - 1;
12     let colBegin = 0;
13     let colEnd = matrix[0].length - 1;
14 
15     while (colBegin <= colEnd && rowBegin <= rowEnd) {
16         // right
17         for (var i = colBegin; i <= colEnd; i++) {
18             res.push(matrix[rowBegin][i]);
19         }
20         rowBegin++;
21 
22         // down
23         for (var i = rowBegin; i <= rowEnd; i++) {
24             res.push(matrix[i][colEnd]);
25         }
26         colEnd--;
27 
28         // left
29         if (rowBegin <= rowEnd) {
30             for (var i = colEnd; i >= colBegin; i--) {
31                 res.push(matrix[rowEnd][i]);
32             }
33         }
34         rowEnd--;
35 
36         // up
37         if (colBegin <= colEnd) {
38             for (var i = rowEnd; i >= rowBegin; i--) {
39                 res.push(matrix[i][colBegin]);
40             }
41         }
42         colBegin++;
43     }
44     return res;
45 };

 

Java实现

 1 class Solution {
 2     public List<Integer> spiralOrder(int[][] matrix) {
 3         List<Integer> res = new ArrayList<>();
 4         // corner case
 5         if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) {
 6             return res;
 7         }
 8 
 9         // normal case
10         int rowBegin = 0;
11         int rowEnd = matrix.length - 1;
12         int colBegin = 0;
13         int colEnd = matrix[0].length - 1;
14         while (rowBegin <= rowEnd && colBegin <= colEnd) {
15             // right
16             for (int i = colBegin; i <= colEnd; i++) {
17                 res.add(matrix[rowBegin][i]);
18             }
19             rowBegin++;
20 
21             // down
22             for (int i = rowBegin; i <= rowEnd; i++) {
23                 res.add(matrix[i][colEnd]);
24             }
25             colEnd--;
26 
27             // left
28             if (rowBegin <= rowEnd) {
29                 for (int i = colEnd; i >= colBegin; i--) {
30                     res.add(matrix[rowEnd][i]);
31                 }
32             }
33             rowEnd--;
34 
35             // up
36             if (colBegin <= colEnd) {
37                 for (int i = rowEnd; i >= rowBegin; i--) {
38                     res.add(matrix[i][colBegin]);
39                 }
40             }
41             colBegin++;
42         }
43         return res;
44     }
45 }

 

LeetCode 题目总结