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