Leetcode 62. Unique Paths dp动态规划，递归会超时

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?

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

Note: m and n will be at most 100.

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

题目链接：https://leetcode.com/problems/unique-paths/

思路：刚开始用递归，超时hh，我把递归代码也贴上。后来想起来用二维数组dp，path[i][j]表示到坐标（i，j)的路径数目

有兴趣还可以看看这题的加强版leetcode 63. Unique Paths II，设置了障碍

class Solution {
public:
    int uniquePaths(int m, int n) {
         vector > path(m+1, vector (n+1, 1));
        for (int i = 2; i <= m; i++)
            for (int j = 2; j <= n; j++)
                path[i][j] = path[i - 1][j] + path[i][j - 1];
        return path[m ][n ];
    }
};

 递归超时的代码：

class Solution {
public:
    void fun(int x,int y,int m,int n,int& res)
    {
        if(x==m||y==n)
        {
            res++;
        }
        else
        {
            if(x