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：

	6	3	1
4	3	2	1
1	1	1	1

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;
        }
    };
}