一道找规律的题
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] ]
可以看成是向右滚了一下矩阵
思路主要是遍历，矩阵的外层到内层滚动数字，每次移动4个数字。
只要想明白了，这个题还是很简单的。
class Solution {public:void rotate(vector<vector<int>>& matrix) {int size = matrix.size();if (size == 0 || size == 1) return;digui(matrix, size, 0);}void digui(vector<vector<int>>& matrix, int size,int i){if (size == 1)return;int zq = size - 1;for (int j = i; j < zq+i; j++){int value = matrix[i][j];matrix[i][j] = matrix[i + i + zq - j][i];//topmatrix[i + zq + i - j][i] = matrix[i + zq][i + zq + i - j];//leftmatrix[i + zq][i + zq + i - j] = matrix[j][i + zq];//bottommatrix[j][i + zq] = value;//right}if (size == 2) return;digui(matrix, size-2, i + 1);}};