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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

解法很简单，以start的坐标为（m,n） finish的坐标为(0,0)

 a[m,n]表示从（m,n）到 (0,0) 的unique paths

从(m,n)走一步会使坐标变为(m-1,n)或(m,n-1)

也就是说a[m,n] = a[m-1,n] + a[m,n-1]

当m或n为1是，路径只有一条

将以上逻辑转为代码就可以了，如下：

    int uniquePaths(int m, int n) {        if(m == 1||n == 1) {            return 1;        }        int **a;        a = new int*[m + 1];        for(int i = 1; i < m + 1; ++i) {            a[i] = new int[n+1];        }        for(int i = 1; i < m+1; ++i) {            for (int j = 1; j < n+1; ++j) {                if(i == 1||j == 1){                    a[i][j] = 1;                }                else{                    a[i][j] = a[i-1][j] + a[i][j-1];                }            }        }        return a[m][n];