ou are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Note: 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 :
Given input matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],
rotate the input matrix in-place such that it becomes:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
Given input matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
rotate the input matrix in-place such that it becomes:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
这个乍一看觉得不难,但是写的时候又不知道怎么回事,其实旋转,对于我们写程序来说,其实就是不停的调换位置,但是怎么调换是个问题。
观察发现,第一个矩阵,最角上的四个1,3,7,9。转完之后,还是这四个数字,只不过是位置变了,接下来这样的四个是:2,4,6,8.最后一个5.再看一下4x4的其实也差不多。
所以想法就是直接每次四个数字进行换,换三次,就能换回来,然后进行下一次调换。 代码如下:
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
if(matrix.size()<=0)
return;
int a=0,b=matrix.size()-1;
while(a<b){
for(int i=0;i<b-a;++i){
swap(matrix[a][a+i],matrix[a+i][b]);
swap(matrix[a][a+i],matrix[b][b-i]);
swap(matrix[a][a+i],matrix[b-i][a]);
}
++a;
--b;
}
}
};