[leetcode]第11周作业

题目：62. Unique Paths

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?

分析：

可以类比那道Climbing Stairs 爬梯子问题，而这道题是每次可以向下走或者向右走，求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样，我们需要用动态规划Dynamic Programming来解。
我们可以维护一个二维数组dp，其中dp[i][j]表示到当前位置不同的走法的个数，然后可以得到递推式为:

dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

代码：

class Solution {public:    int uniquePaths(int m, int n) {        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];    }};