[Leetcode] Rotate Image

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

用了最笨的办法。就地转的话应该是一圈一圈的转吧，有空再写写。

 1 class Solution {

 2 public:

 3     void rotate(vector<vector<int> > &matrix) {

 4         int n = matrix.size();

 5         vector<vector<int> > tmp(n, vector<int>(n));

 6         for (int i = 0; i < n; ++i) {

 7             for (int j = 0; j < n; ++j) {

 8                 tmp[i][j] = matrix[i][j];

 9             }

10         }

11         for (int i = 0; i < n; ++i) {

12             for (int j = 0; j < matrix[0].size(); ++j) {

13                 matrix[j][n - 1 - i] = tmp[i][j];

14             }

15         }

16     }

17 };

第二次刷leetcode发现好多题第一次做时就是A掉就不管了，其实还有很多东西可以继续挖掘，所以！要多刷几遍啊！原地转的话，这里用到一个小技巧，把一张纸上下折，再按对角线对折，那么它就转了90度了有木有，所以这里就用了这个技巧。

 1 class Solution {

 2 public:

 3     void rotate(vector<vector<int> > &matrix) {

 4         if (matrix.empty()) return;

 5         int n = matrix.size(), tmp;

 6         for (int i = 0; i < n/2; ++i) {

 7             for (int j = 0; j < n; ++j) {

 8                 tmp = matrix[i][j];

 9                 matrix[i][j] = matrix[n-1-i][j];

10                 matrix[n-1-i][j] = tmp;

11             }

12         }

13         for (int i = 0; i < n; ++i) {

14             for (int j = 0; j < i; ++j) {

15                 tmp = matrix[i][j];

16                 matrix[i][j] = matrix[j][i];

17                 matrix[j][i] = tmp;

18             }

19         }

20         return;

21     }

22 };