AI代码解释 classSolution:defrotate(self,matrix:List[List[int]])->None:""" Do notreturnanything,modify matrixin-place instead.""" n=len(matrix)# 注意一下范围和下标即可foriinrange(n//2):forjinrange(i,n-1-i):matrix[i][j],matrix[j][n-1-i],matrix[n-1-i][n-1-j],matrix[n-1...
publicvoidrotate(int[][] matrix) {intn =matrix.length;int[][] m =newint[n][n];for(introw=0;row<n;row++){for(intcol=0;col<n;col++){ m[row][col]= matrix[n-1-col][row]; } }//再赋值回matrix,注意java是形参是引用传递for(introw=0;row<n;row++){for(intcol=0;col<n;col...
classSolution{public:voidrotate(vector<vector<int>>&matrix){int rownum=matrix.size();int colnum=matrix[0].size();// 将矩阵转置for(int i=0;i<rownum;i++){for(int j=0;j<i;j++){swap(matrix[i][j],matrix[j][i]);}}// 每一行对称翻转for(int i=0;i<rownum;i++){reverse(matrix[...
{for(intj =0; j < N/2; ++j )// 次数确定:每次减两个,直到只剩下一个或者零个swap( matrix[i][j] , matrix[i][N-j-1] ); }// printPrintMatrix( matrix,"After rotate 90 degree\n"); }voidRotate90Degree(intmatrix[N][N] ){// printPrintMatrix( matrix,"Before");for(inti =0; ...
我们再重复一次之前的操作, matrix[n - row - 1][n - col - 1]经过旋转之后会到哪一个位置上呢? 自己看吧???https://leetcode.cn/problems/rotate-image/solution/xuan-zhuan-tu-xiang-by-leetcode-solution-vu3m/ class Solution: def rotate(self, matrix: List[List[int]]) -> None: n...
class Solution { public: void rotate(vector<vector<int>>& matrix) { int n = matrix.size(); for (int i = 0; i < n / 2; ++i) { for (int j = 0; j < (n + 1) / 2; ++j) { int temp = matrix[i][j]; matrix[i][j] = matrix[n - j - 1][i]; matrix[n - j ...
Rotate Image 二维数组旋转90度 Rotate Image You are given an n x n Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? class Solution { public: void rotate(vector<vector<int> > &matrix) { //方法1:tempMatrix[j][n-1-i] = matrix[i][j]; 置换...
给定matrix = [ [1,2,3], [4,5,6], [7,8,9] ], 原地旋转输入矩阵,使其变为: [ [7,4,1], [8,5,2], [9,6,3] ] 代码 class Solution { public: void rotate(vector<vector<int>>& matrix) { // 先沿对角线翻转 int n = matrix.size(); for(int i = 0; i < n; i ++)...
Solution2：翻转+对角线互换class Solution { public: void rotate(vector<vector<int>>& matrix)...
matrix[i].reverse()#然后将矩阵的每一行翻转returnmatrix 解题思路2:首先沿着副对角线翻转一次,然后沿着水平中线翻转一次,就可以得到所要求的矩阵了。时间复杂度O(n^2),空间复杂度O(1) classSolution:#@param matrix, a list of lists of integers#@return a list of lists of integersdefrotate(self, matrix...