Unique Paths(faster than 100.00% and less than 75.54%)

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?
一开始拿到这道题的时候我是用的暴力，果不其然超时了TnT，因此只能改用打表了。
超时代码：

class Solution {
public:
    void isReach(int i, int j, const int &m, const int &n, int &count){
        if(i > m || j > n)return;
        else if(i == m and j == n){
            count++;
            return;
        }
        else{
            isReach(i + 1, j, m, n, count);
            isReach(i, j + 1, m, n, count);
        }
    }
    int uniquePaths(int m, int n) {
        int count = 0;
        isReach(1, 1, m, n, count);
        return count;
    }
};

通过代码：

class Solution {
public:
    int uniquePaths(int m, int n) {
        int dp[101][101] = { 0 };
    for (int i = 0; i < m; i++)
        dp[0][i] = 1;
    for (int j = 1; j < n; j++)
        dp[j][0] = 1;
    for (int i = 1; i < n; i++)
        for (int j = 1; j < m; j++)
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];   
    return dp[n-1][m-1];
    }
};

本文地址：https://blog.csdn.net/i_actor/article/details/110190526