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?

不使用额外的空间顺时针旋转方阵90度

例如

旋转后变为

算法1

先将矩阵转置，然后把转置后的矩阵每一行翻转

转置变为每一行翻转变为

class Solution {

public:

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

int n = matrix.size();

//转置

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

for(int j = i+1; j < n; j++)

swap(matrix[i][j] , matrix[j][i]);

//每一行翻转

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

for(int j = 0; j < (n>>1); j++)

swap(matrix[i][j], matrix[i][n-j-1]);

}

};

算法2

可以见矩阵看成多个环组成，如下4*4的矩阵包括两个环，第一个环为1,2,3,4,8,12,16,15,14,13,9,5,1，第二个环为6,7,11,10。

旋转一个矩阵，相当于把每一个环都旋转。如何旋转一个环呢？以最外层的环举例：本文地址

旋转前：，旋转后：

我们把环分成3组：{1,4,16,13}，{2,8,15,9}，{3,12,14,5}，每一组中：旋转后相当于把原来的数字移动到同组中下一个数字的位置

对于一个n*n的矩阵可以分成n/2（向上取整）个环来旋转；对于边长为len的环，可以分成len-1组来旋转。

class Solution {

public:

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

int n = matrix.size();

if(n == 0)return;

for(int i = 0, len = n; i < n/2; i++, len -= 2)

{//n/2 为旋转的圈数，len为第i圈中正方形的边长

int m = len - 1;

for(int j = 0; j < m; j++)

{

int tmp = matrix[i][i+j];

matrix[i][i+j] = matrix[i+m-j][i];

matrix[i+m-j][i] = matrix[i+m][i+m-j];

matrix[i+m][i+m-j] = matrix[i+j][i+m];

matrix[i+j][i+m] = tmp;

}

};

本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3768734.html，如需转载请自行联系原作者

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