原题链接在这里:https://leetcode.com/problems/unique-paths-iii/ 题目: On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending s…

On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 repre…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 回溯法 日期 题目地址:https://leetcode.com/problems/unique-paths-iii/ 题目描述 On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square. There is exactly one s…

题目来源: https://leetcode.com/problems/unique-paths-iii/ 自我感觉难度/真实难度: 题意: 分析: 回溯法,直接DFS就可以了 自己的代码: class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: res=0 n=len(grid) m=len(grid[0]) for i in range(n): for j in range(m): if grid[i][…

On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 repre…

题目如下: On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1…

Leetcode之深度优先搜索&回溯专题-980. 不同路径 III(Unique Paths III) 深度优先搜索的解题详细介绍,点击 在二维网格 grid 上,有 4 种类型的方格: 1 表示起始方格.且只有一个起始方格. 2 表示结束方格,且只有一个结束方格. 0 表示我们可以走过的空方格. -1 表示我们无法跨越的障碍. 返回在四个方向(上.下.左.右)上行走时,从起始方格到结束方格的不同路径的数目,每一个无障碍方格都要通过一次. 示例 1: 输入:[[1,0,0,0],[0,0,0,…

On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 repre…

原题链接 字母题 : unique paths Ⅱ 思路: dp[i][j]保存走到第i,j格共有几种走法. 因为只能走→或者↓,所以边界条件dp[0][j]+=dp[0][j-1] 同时容易得出递推 dp[i][j]+=dp[i-1][j]+dp[i][j-1] class Solution { public: int uniquePaths(int m, int n) { if (m == 0 || n == 0) { return 0; } vector<vector<int>&g…

转载请注明原文地址:http://www.cnblogs.com/LadyLex/p/9057297.html 原题链接: 今天考试考了前天的SDOI考题 天啊我菜爆,只有T2拿了30分 然后考试后半程一直在打T1 觉得考试思路很有意思,于是就顺着打下来了 个人感觉这个是$O(nlog^{2}n)$的,但是在loj上我比claris的程序快了1s多,只不过编程复杂度不止翻倍啊…… 下面介绍一下我的解法 其实最早启发我的是链上的部分分 定义$pre_{i}$为i前面最近的和i同色的点的下标,我们把…