leetcode 62. Unique Paths DP动态规划

我不是女神ヾ 2022-06-09 11:39 156阅读 0赞

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?  
![这里写图片描述][SouthEast]  
Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

这道题求解的是从开始到结束点的所有路径的数量，可以使用动态规划了来做，也可以使用递归来做。 这里使用dp\[i\]\[j\]的值即为路径的数量  
转移方程是 dp\[i\]\[j\] = dp\[i-1\]\[j\] + dp\[i\]\[j-1\]  
边界条件是 dp\[i\]\[1\] = 1, dp\[1\]\[j\] = 1

其实可以发现这是一个斐波那契数列，嗯嗯，就是这么简单，但是包装起来一点都不想斐波那契数列。

代码如下：

public class Solution
    {
        public int uniquePaths(int m, int n)
        {
            if(m<=0 || n<=0)
                return 0;
            //return getByRecursion(m, n);
            return byDP(m, n);
        }
    
        /*
         * dp[i][j]的值即为路径的数量 
         * dp[i][j] = dp[i-1][j] + dp[i][j-1]
         * 边界条件：dp[i][1] = 1, dp[1][j] = 1
         * */
        public int byDP(int m, int n)
        {   
            int [][] mat=new int[m][n];
            for(int i=0;i<m;i++)
                mat[i][0]=1;
            for(int j=0;j<n;j++)
                mat[0][j]=1;
    
            for(int i=1;i<m;i++)
            {
                for(int j=1;j<n;j++)
                    mat[i][j]=mat[i-1][j]+mat[i][j-1];
            }
            return mat[m-1][n-1];
        }
    
        /*
         * 这个问题的递归算法类似斐波拉契数列
         * 这个递归结构也就告诉了我们的动态规划的转换方程
         * 但是会超时
         * */
         public int getByRecursion(int m, int n)
         {
                if(m==1 && n==1)
                    return 1;
                else 
                    return uniquePaths(m-1, n)+uniquePaths(m, n-1);
         }
    }

下面是C++的做法，就是一个普通的DP

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution 
    {
    public:
        int uniquePaths(int m, int n) 
        {
            vector<vector<int>> dp(m, vector<int>(n, 0));
            for (int i = 0; i < m; i++)
                dp[i][0] = 1;
            for (int i = 0; i < n; i++)
                dp[0][i] = 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];
        }
    };

[SouthEast]: /images/20220609/50bb32a0dd054561bef1c54ad43d4c27.png