https://leetcode.cn/problems/rotate-image/

👍逐圈遍历

这个还挺符合自己想法的，思路差不多，但是没写出来。 参考答案：https://leetcode.cn/problems/rotate-image/solution/li-kou-48xiao-bai-du-neng-kan-dong-de-fang-fa-zhu-/ 参考了答案后写出来了，这个还是最好理解的，每圈的左上角坐标为（pos1，pos1），右下角坐标为（pos2，pos2），然后逐圈遍历。

  1. public void rotate(int[][] matrix) {
  2.     int pos1 = 0, pos2 = matrix.length - 1;
  3.     while (pos1 < pos2) {
  4.         int add = 0;
  5.         while (pos1 + add < pos2) {
  6.             int temp = matrix[pos1][pos1 + add];
  7.             matrix[pos1][pos1 + add] = matrix[pos2 - add][pos1];
  8.             matrix[pos2 - add][pos1] = matrix[pos2][pos2 - add];
  9.             matrix[pos2][pos2 - add] = matrix[pos1 + add][pos2];
  10.             matrix[pos1 + add][pos2] = temp;
  11.             add++;
  12.         }
  13.         pos1++;
  14.         pos2--;
  15.     }
  16. }

分组旋转

不能借助辅助数组还挺难的，想不到… 参考答案：https://leetcode.cn/problems/rotate-image/solution/48-xuan-zhuan-tu-xiang-fu-zhu-ju-zhen-yu-jobi/

  1. public void rotate(int[][] matrix) {
  2.     int n = matrix.length;
  3.     for (int i = 0; i < n / 2; i++) {
  4.         for (int j = 0; j < (n + 1) / 2; j++) {
  5.             int tmp = matrix[i][j];
  6.             matrix[i][j] = matrix[n - 1 - j][i];
  7.             matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
  8.             matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
  9.             matrix[j][n - 1 - i] = tmp;
  10.         }
  11.     }
  12. }

巧解

“对角线反转+逐行反转” 参考答案：https://leetcode.cn/problems/rotate-image/solution/-by-huan-huan-20-8df6/ 太巧妙了，我自己是想不到…😥