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

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

Constraints:

• 1 <= m, n <= 100
• It’s guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.

---

问从左上角走到右下角一共有多少种方案，每次只能往右或者下走。很简单的题，高中学过排列组合的都知道。 假设右下和左上的坐标之差为(x,y)，则从左上走到右下一共需要走x+y步，可以分为x步往右走，y步往下走。每一步都可以选择往右或者往下走，所以总的方案数为$C_{x+y}^x==C_{x+y}^y$，用x,y中的较小者来算就可以了，计算方法就是高中数学老师教的啦:-) 完整代码如下

class Solution {
public:
    int uniquePaths(int m, int n)
    {
        int x = m – 1, y = n – 1;
        int z = x + y, u = x < y ? x : y;
        double ans = 1.0;
        for (int i = 1; i <= u; i++) {
            ans *= z;
            ans /= i;
            z–;
        }
        return ans;
    }
};

本代码提交AC，用时0MS。