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No.272 - LeetCode[861] Score After Flipping Matrix - 贪心
贪心 /* * @lc app=leetcode id=861 lang=cpp * * [861] Score After Flipping Matrix */ // @lc code=start class Solution { public: int calculate(vector<vector<int>>& grid) { int N = grid.size(); int M = grid[0].size();Leetcode 861. 翻转矩阵后的得分（DAY 117) ---- 贪心算法学习期
原题题目 代码实现（首刷自解） class Solution { public: int matrixScore(vector<vector<int>>& grid) { int ret = 0; for(int x=0;x<grid.size();++x) { if(grid[x][0]) continue; else {AcWing 861. 二分图的最大匹配
二分图的最大匹配 ①. 题目②. 思路③. 学习点④. 代码实现 原题链接 ①. 题目 ②. 思路 ③. 学习点 ④. 代码实现861. Score After Flipping Matrix(Leetcode每日一题-2020.12.07)
Problem We have a two dimensional matrix A where each value is 0 or 1. A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s. After making any number of moves, every row of t力扣日常#861 翻转矩阵后的得分
1 class Solution { 2 public int matrixScore(int[][] A) { 3 int len1 = A.length, len2 = A[0].length; 4 5 for(int i = 0; i < len1; i++){ 6 if(A[i][0] == 0){ 7 for(int j = 0; j < len2; j++){ 8Leetcode之861. 翻转矩阵后的得分（题解）
题目：有一个二维矩阵 A 其中每个元素的值为 0 或 1 。 移动是指选择任一行或列，并转换该行或列中的每一个值：将所有 0 都更改为 1，将所有 1 都更改为 0。 在做出任意次数的移动后，将该矩阵的每一行都按照二进制数来解释，矩阵的得分就是这些数字的总和。 返回尽可能高的分数。 示例[leetcode] 861. Score After Flipping Matrix
Description We have a two dimensional matrix A where each value is 0 or 1. A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s. After making any number of moves, every row