62. 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.

因为机器人只能向右或者向下走，所以当前点可以到达finish的路径数等于右点和下点可以到达finish的路径和。以3*4为例，步数如下表：

 63143211111填的时候从右下角开始，先写一，然后逐次填左和上，有点像扫雷的数字计算规则。根据这个思路，代码如下：

public class Solution {    public int uniquePaths(int m, int n) {        if (m == 0 || n == 0) {            return 0;        }        int[][] res = new int[m][n];        for (int i = 0; i < m; i ++) {            res[i][n - 1] = 1;        }        for (int i = 0; i < n; i ++) {            res[m - 1][i] = 1;        }        for (int i = m - 2; i >= 0; i --) {            for (int j = n - 2; j >= 0; j --) {                res[i][j] = res[i + 1][j] + res[i][j + 1];            }        }        return res[0][0];    }}

还有另一种思路，按照排列组合的思想，总的要走的步数N是 m+n-2，需要向下走的步数K是 m-1，那么可能的选择有 C N K，一个公式搞定，代码如下：

class Solution {    public:        int uniquePaths(int m, int n) {            int N = n + m - 2;// how much steps we need to do            int k = m - 1; // number of steps that need to go down            double res = 1;            // here we calculate the total possible path number             // Combination(N, k) = n! / (k!(n - k)!)            // reduce the numerator and denominator and get            // C = ( (n - k + 1) * (n - k + 2) * ... * n ) / k!            for (int i = 1; i <= k; i++)                res = res * (N - k + i) / i;            return (int)res;        }    };}