Unique Paths

Description
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：题目要求算出从start到finish位置一共有多少种可能的走法，只能往右或往下走的限制带来了很大的方便，也就是说相对现在所处的位置，上一个位置在当前位置的左边或右边，因此用动态规划的方法就能很容易地解决问题，dp（i，j）表示从起点走到坐标为（i，j）的点的方法数，状态方程为dp（i，j）=dp（i-1，j）+dp（i，j-1），由于走到第一行中任意一个位置的走法都只能一种，就是一直往右，走到第一列中的任意一个位置也一样，一直往下，所以算出到达这些位置的走法以后，就能够一直递推下去了。程序代码如下：

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