Unique Paths

题目描述：

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

Note: m and n will be at most 100.

解题思路：

使用动态规划,dp[i][j]表示从起点（0，0）到（i,j）的不同路径的数目，由于只能向右和向下移动，

所以状态转移方程为：dp[i][j]=dp[i-1][j]+dp[i][j-1]，最后的dp[m-1][n-1]即为所求。

AC代码如下：

class Solution {
public:
int uniquePaths(int m, int n) {
if (m <= 0 || n <= 0) return 0;
vector<vector<int>> dp(m, vector<int>(n, 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];
}
};