062 Unique Paths 不同路径】的更多相关文章

机器人位于一个 m x n 网格的左上角, 在下图中标记为“Start” (开始).机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角,在下图中标记为“Finish”(结束).问有多少条不同的路径? 详见:https://leetcode.com/problems/unique-paths/description/ Java实现: 参考:https://www.cnblogs.com/springfor/p/3886603.html class Solution { public i…
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…
题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish'…
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…
题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish'…
一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为"Start" ). 机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角(在下图中标记为"Finish"). 问总共有多少条不同的路径? 例如,上图是一个7 x 3 的网格.有多少可能的路径? 说明:m 和 n 的值均不超过 100. 示例 1: 输入: m = 3, n = 2 输出: 3 解释: 从左上角开始,总共有 3 条路径可以到达右下角. 1. 向右 -> 向右 ->…
网址:https://leetcode.com/problems/unique-paths/ 第一思路是动态规划 通过观察,每一个格子的路线数等于相邻的左方格子的路线数加上上方格子的路线数 于是我们就得到 dp[i][j] = dp[i-1][j] + dp[i][j-1] 这个状态转移方程 一开始通过调用函数的方式对dp数组赋值,并且每次都判断index是否为特殊值,感觉耗时会很多,果不其然! int get_sum(vector<vector<int>> dp, int i,…
题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish'…