zoukankan      html  css  js  c++  java
  • 62. Unique Paths (Graph; DP)

    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 the diagram below).

    How many possible unique paths are there?

    思路: 如果使用递归,那么DFS(i,j)=v[i][j]+max{DFS(i,j-1), DFS(i-1,j)},问题是DFS(i,j)可能会被计算两次,一次来自它右边的节点,一次来自它下面的节点。每个节点都被计算两次,时间复杂度就是指数级的了。解决方法是使用动态规划存储节点信息,避免重复计算。

    Above is a 3 x 7 grid. How many possible unique paths are there?

    Note: m and n will be at most 100.

    class Solution {
    public:
        int uniquePaths(int m, int n) {
            int dp[m][n];
            
            dp[0][0] = 1;
            for(int i = 0; i< n; i++ )
            {
                dp[0][i] = 1;
            }
            for(int i = 0; i< m; i++ )
            {
                dp[i][0] = 1;
            }
            
            for(int i = 1; i< m; i++)
            {
                for(int j = 1; j< n; j++)
                {
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
            return dp[m-1][n-1];
        }
    };
  • 相关阅读:
    iOS—UI —推送实现
    iOS—UI —懒加载
    iOS多线程和NSRunLoop概述
    ios安全性---AES加密
    iOS私有API
    iOS多线程 && Runloop
    iOS毛玻璃效果
    Swift -4-对象与类
    Swift -3-函数&闭包
    Swift -1- 简介&简单值&基本类型
  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/4854804.html
Copyright © 2011-2022 走看看