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.

题目解析：

让找从起始点出发一共有多少路径到达终点。这也就是动态规划问题。到达(i,j)点，只可能从(i-1,j) (i,j-1)到达。因此arr[i][j] = arr[i-1][j] + arr[i][j-1]。

有了表达式我们可以看出只需要一个一维数组，就能将问题解决。arr[i] = arr[i-1]+arr[i]。节省了很大的空间。

class Solution {
public:
    int uniquePaths(int m, int n) {
        int *arr = new int[n];

        for(int i = 0;i < n;i++)
            arr[i] = 1;

        for(int i = 1;i < m;i++){
            for(int j = 1;j < n;j++)
                arr[j] = arr[j-1]+arr[j];
        }
        int res = arr[n-1];
        delete []arr;
        return res;
    }
};