【Leetcode_easy】867. Transpose Matrix

problem 766. Toeplitz Matrix solution1: class Solution { public: bool isToeplitzMatrix(vector<vector<int>>& matrix) { for(int i=0; i<matrix.size()-1; ++i) { for(int j=0; j<matrix[0].size()-1; ++j) { if(matrix[i][j] != matrix[i+1][j+1

http://poj.org/problem?id=3422 (题目链接) 题意 N*N的方格,每个格子中有一个数,寻找从(1,1)走到(N,N)的K条路径,使得取到的数的和最大. Solution 同[codevs1277] 方格取数 代码 // poj3422 #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio>

Question 867. Transpose Matrix Solution 题目大意:矩阵的转置 思路:定义一个转置后的二维数组,遍历原数组,在赋值时行号列号互换即可 Java实现: public int[][] transpose(int[][] A) { int[][] B = new int[A[0].length][A.length]; for (int i = 0; i < A.length; i++) { for (int j = 0; j < A[0].length; j++

矩阵转置,A[i][j] 变成A[j][i] 比较简单,直接上代码了. func transpose(A [][]int) [][]int { B := make([][]int, len(A[0])) for i := 0; i < len(A[0]); i++ { B[i] = make([]int, len(A)) for j := 0; j < len(A); j++ { B[i][j] = A[j][i] } } return B } 原文地址:https://blog.51cto.

[官网]https://docs.scipy.org/doc/numpy-1.15.0/reference/generated/numpy.transpose.html numpy.transpose(a, axes=None)[source] 参数a:array_like,输入数组. 参数axes:int列表,可选,默认情况下,反转维度,否则根据给定的值排列坐标轴. Examples >>> x = np.arange(4).reshape((2,2)) >>> x

题目要求 Given a matrix A, return the transpose of A. The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix. 题目分析及思路 题目要求得到矩阵的转置矩阵.可先得到一个行数与原矩阵列数相等.列数与原矩阵行数相等的矩阵,再对原矩阵进行遍历. python代码? c

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right.The first integer of each row is greater than the last integer of the previous row

Problem description Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. The distance between two adjacent cells is 1. Example 1: Input: 0 0 0 0 1 0 0 0 0 Output: 0 0 0 0 1 0 0 0 0 Example 2: Input: 0 0 0 0 1 0 1 1 1

问题描述:将二维数组中值为0的元素,所在行或者列全set为0:https://leetcode.com/problems/set-matrix-zeroes/ 问题分析:题中要求用 constant space 的辅助空间.自然想到位优化.一个int可以存储31个元素的信息.这里刚好用到了字符串论文里面常用的优化处理方法.类似桶分的思想.好吧,这么看来这长时间的论文没白看. 附上代码: 1 void setZeroes(vector<vector<int>>& matrix