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?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
典型的一个动态规划的题目。思路比较简单，当前格子的路径数等于当前格子左一格子的路径数加上当前格子上一个格子的路径数。注意的是边界情况。如果行列有为1的话 直接返回。否则下面的处理会越界。
贴上代码：
class Solution { public: int uniquePaths(int m, int n) { int grid[m][n]; grid[0][0] = 0; if(m == 1 || n == 1){return 1;} for(int j = 1; j < n; ++j){ grid[0][j] = 1; } for(int i = 1; i < m; ++i){ grid[i][0] = 1; } for(int i = 1; i < m; ++i){ for(int j = 1; j < n; ++j){ grid[i][j] = grid[i-1][j]+grid[i][j-1]; } } return grid[m-1][n-1]; } };