动态规划----unique paths

1、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' in the diagram below).

How many possible unique paths are there?

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

2、解答： f(n,m)只能从上或者左边走过来，因此f(n,m) = f(n-1,m) + f(n,m-1)。现在看起始条件，若n = 0 ,f(0,m) = 1,m=0,f(n,0) = 1

3、代码：

C++代码

    

class Solution {
public:
    int uniquePaths(int m, int n) {
        int path[m][n]; 
        
        for(int i=0;i<m;i++)
            path[i][0] = 1;
        
        for(int j=0;j<n;j++)
            path[0][j] = 1;
        
        for(int i=1;i<m;i++)
            for(int j=1;j<n;j++)
                path[i][j] = path[i-1][j] + path[i][j-1];
        
        return path[m-1][n-1];
    } 
};

python代码

class Solution:
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        path = [[1 for i in range(n)] for j in range(m)]
       
        for i in range(1,m):
            for j in range(1,n):
                path[i][j] = path[i-1][j] + path[i][j-1]
        
        return path[-1][-1]