算法第14周Unique Paths[medium]

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?
Note: m and n will be at most 100.

---

Solution

对于任意位置来说，robot仅可能从它左边的格子或者上面的格子来。
我们采用动态规划的方法：
dp[i][j] 代表达到（i,j）的路径个数。
我们可以知道dp[i][j] = dp[i-1][j]+dp[i][j-1];
我们必须考虑边界的情况。
当位于第0行或第0列时，路径个数恒为1；

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