[LeetCode] 48. Rotate Image

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

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

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Example 3:

Input: matrix = [[1]]
Output: [[1]]

Example 4:

Input: matrix = [[1,2],[3,4]]
Output: [[3,1],[4,2]]

Constraints:

matrix.length == n
matrix[i].length == n
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000

旋转图像。

题意是给一个n x n的二维数组，请你将它顺时针九十度旋转。要求in-place做。

既然规定了要 in-place 做，只能想到对折之类的做法。但是如何对折呢？是先通过对角线对折（左上 - 右下），然后再上下对折。注意对角线对折部分 j 指针是从哪里开始的。

时间O(n^2)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {number[][]} matrix
 3  * @return {void} Do not return anything, modify matrix in-place instead.
 4  */
 5 var rotate = function (matrix) {
 6     let m = matrix.length;
 7     // 左上- 右下对角线交换
 8     for (let i = 0; i < m; i++) {
 9         for (let j = i; j < m; j++) {
10             let temp = matrix[i][j];
11             matrix[i][j] = matrix[j][i];
12             matrix[j][i] = temp;
13         }
14     }
15 
16     // 左右对折
17     for (let i = 0; i < m; i++) {
18         for (let j = 0; j < m / 2; j++) {
19             let temp = matrix[i][j];
20             matrix[i][j] = matrix[i][m - 1 - j];
21             matrix[i][m - 1 - j] = temp;
22         }
23     }
24 };

Java实现

 1 class Solution {
 2     public void rotate(int[][] matrix) {
 3         // 按左上 - 右下对角线交换
 4         int m = matrix.length;
 5         for (int i = 0; i < m; i++) {
 6             for (int j = i; j < m; j++) {
 7                 int temp = matrix[i][j];
 8                 matrix[i][j] = matrix[j][i];
 9                 matrix[j][i] = temp;
10             }
11         }
12 
13         // 左右对折
14         for (int i = 0; i < m; i++) {
15             for (int j = 0; j < m / 2; j++) {
16                 int temp = matrix[i][j];
17                 matrix[i][j] = matrix[i][m - 1 - j];
18                 matrix[i][m - 1 - j] = temp;
19             }
20         }
21         return;
22     }
23 }