You 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 1:
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]
]
Example 2:
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]
]
题目大意：
给定一个n x n的矩阵，要求不借助额外的存储空间，将矩阵顺时针旋转90度。
解法：
脑袋里面没有想到什么解法，研究了数组的下标规则，也没有想出太多。
参考了网上大佬们的解法。真是神仙解法。
先将矩阵上下翻转，再将矩阵对角线翻转。
first reverse up to down, then swap the symmetry
* 1 2 3 7 8 9 7 4 1
* 4 5 6 => 4 5 6 => 8 5 2
* 7 8 9 1 2 3 9 6 3
C++：class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
reverse(matrix.begin(),matrix.end());
for(int i=0;i<matrix.size();i++){
for(int j=i+1;j<matrix.size();j++){
swap(matrix[i][j],matrix[j][i]);
}
}
}
};
Python:
class Solution(object):
def rotate(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: None Do not return anything, modify matrix in-place instead.
"""
matrix.reverse()
for i in range(0,len(matrix)):
for j in range(i+1,len(matrix)):
matrix[i][j],matrix[j][i]=matrix[j][i],matrix[i][j]