[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 };
原文地址:https://www.cnblogs.com/easonliu/p/3649422.html