matrix[i], matrix[n - 1 - i] = matrix[n - 1 - i], matrix[i] for i in range(1,n): #Begin with '1' instead of 0 can avoide action on main-diagonal for j in range(i): #沿着主对角线翻转; matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] return matrix""...
privateint[][] helper(int[][] matrix) { // 按左上 - 右下对角线交换 intm=matrix.length; for(inti=0; i < m; i++) { for(intj=i; j < m; j++) { inttemp=matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } // 左右对折 for(inti=0; i < m;...
Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 (e.g.,"waterbottle" is a rotation of"erbottlewat"). Can you use only one call to the method that checks if one word is a substring of another? Example 1: Input: s1 = "waterbottle", s2 = "erbottl...
Given an array of integers A and let n to be its length. 15220 LeetCode 0048 - Rotate Imageimageinputmatrixrotation Reck Zhang 2021-08-11 You are given an n x n 2D matrix representing an image. 42320 PDF转图片bugpdfpiprotationtraits GhostCN_Z 2021-08-04 经过测试,fitz是目前PDF转图片...
y2=matrix[0].size()-1; while(x1<=x2&&y1<=y2) { for(i=y1;i<=y2;i++) res.push_back(matrix[x1][i]); for(i=x1+1;i<=x2;i++) res.push_back(matrix[i][y2]); if(x2!=x1) for(i=y2-1;i>=y1;i--) res.push_back(matrix[x2][i]); ...