LeetCode62. Unique Paths（动态规划）

Medium
Accepted：270,453
Submissions：577,298

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

链接

https://leetcode.com/problems/unique-paths

题意

一个机器人位于一个m*n方格的左上角，现在它要走到右下角，每次可以向右或者向下走一格，问有多种不同的路径。

题解

之前我们做的题是求一条从左上角到右下角最短的路径，LeetCode64. Minimum Path Sum（动态规划），这道题其实思路也差不多。

考虑当前某个格子，只能从上边和左边到达。设$f[i][j]$表示从左上角到达方格（i, j）的路径树，那么$f[i][j] = f[i-1][j] + f[i][j-1]$，实际上我们每次按行更新，所以可以只使用一个一维数组。

时间复杂度$O(nm)$，空间复杂度$O(m)$。

当然还可以考虑使用数学方法做。

代码

Runtime: 8 ms, faster than 11.52% of C++ online submissions for Unique Paths.
Memory Usage: 8.6 MB, less than 45.22% of C++ online submissions for Unique Paths.

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector <int> v(m + 1, 1);
        for (int i = 2; i <= n; ++i)
        {
            for (int j = 2; j <= m; ++j)
                v[j] = v[j - 1] + v[j];
        }
        return v[m];
    }
};

The end.
2019年4月1日星期一