［抄题］: 求挖掉一些区域后,能允许出现的最大十字架 In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus s…

[LeetCode]764. Largest Plus Sign 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/largest-plus-sign/description/ 题目描述: In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except thos…

根据题意的话就是在非0的地方开始寻找上下左右分别能够走到的最大步长的. 那么使用暴力的方法竟然leetcode还是给过了. class Solution { public: int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { ; vector<vector<)); ;i<mines.size();i++){ vis[mines[i][]][mines[i][]]=; } ;i<…

题目大意: 就是一个由1和0组成的正方形矩阵,求里面最大的加号的大小,这个大小就是长度. 什么鬼啊,本来想自己想的,结果看了半天没看懂具体什么意思,然后查了下题解,希望有人说一下意思,结果一上来就是思路,还直接动态规划四个大字,我也是呵呵了 思路一:暴力Brute Force 就是用i, j 循环每一个位置,判断该位置的上下左右的最长“1序列”,复杂度大概是n3 ***这个据说过不了 思路二:动态规划Dynamic Programming 说实话看完源码才想到的,感觉挺好的,直接看源码大把,很好…

In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, retu…

In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, retu…

题目如下: In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none…

题目 A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population. Input Specification: Each input file contains one test…

题意: 输入两个正整数N和M(N<100,M<N),表示结点数量和有孩子结点的结点数量,输出拥有结点最多的层的结点数量和层号(根节点为01,层数为1,层号向下递增). AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; vector<]; ]; void dfs(int x,int storey){ ++ans[storey]; for(aut…

思路: 首先使用dp计算出在每个位置(i, j)上下左右最多有多少个连续的1,得到up[i][j], down[i][j], left[i][j], right[i][j].然后计算这四个值中的最小值,所有最小值中的最大值就是答案. 实现: class Solution { public: int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { vector<vector<)), right…