leetcode 腾讯精选练习（50 题）62.不同路径

原题目

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?

Note: m and n will be at most 100.

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

思路

第一遍解法

网上好的解法

class Solution {
    public:
        int uniquePaths(int m, int n) {
            int N = n + m - 2;// 一共需要走的步数
            int k = m - 1; // 需要向下走的步数
            double res = 1;
            // 计算所有的路径 
            // 当向下的步数和向右的步数不同时，路径 一定 不同
            // 当向下的步数和向右的步数相同时 假设 m = 2, n = 3
            // 向下走 2-1=1 步 向右走 3+2-2-1=2 步 一共走 2+1=3 步
            // 向下走的顺序可以是第一步、第二步、第三步 
            // 计算公式就是 C(N, k) = n! / (k!(n - k)!)
            // 化简之后的计算公式
            // C = ((n-k+1)*(n-k+2)*...*(n-k+k))/1*2*...*k
            //   = (n-k+1)/1 * (n-k+2)/2 *...*(n-k+k)/k
            for (int i = 1; i <= k; i++)
                res = res * (N - k + i) / i;
            return (int)res;
        }
    };

自己可以改进的地方

最简代码

获得的思考