https://leetcode.com/problems/unique-paths/description/
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
这是一道典型的适合使用动态规划解决的题目,它和爬楼梯等都属于动态规划中最简单的题目,因此也经常会被用于面试之中。
读完题目你就能想到动态规划的话,建立模型并解决恐怕不是难事。其实我们很容易看出,由于机器人只能右移动和下移动, 因此第[i, j]个格子的总数应该等于[i - 1, j] + [i, j -1], 因为第[i,j]个格子一定是从左边或者上面移动过来的。
代码大概是:
JS Code:
const dp = [];
for (let i = 0; i < m + 1; i++) {
dp[i] = [];
dp[i][0] = 0;
}
for (let i = 0; i < n + 1; i++) {
dp[0][i] = 0;
}
for (let i = 1; i < m + 1; i++) {
for(let j = 1; j < n + 1; j++) {
dp[i][j] = j === 1 ? 1 : dp[i - 1][j] + dp[i][j - 1]; // 转移方程
}
}
return dp[m][n];
Python Code:
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
d = [[1] * n for _ in range(m)]
for col in range(1, m):
for row in range(1, n):
d[col][row] = d[col - 1][row] + d[col][row - 1]
return d[m - 1][n - 1]
复杂度分析
- 时间复杂度:$O(M * N)$
- 空间复杂度:$O(M * N)$
由于dp[i][j] 只依赖于左边的元素和上面的元素,因此空间复杂度可以进一步优化, 优化到O(n).
具体代码请查看代码区。
当然你也可以使用记忆化递归的方式来进行,由于递归深度的原因,性能比上面的方法差不少:
直接暴力递归的话会超时。
Python3 Code:
class Solution:
visited = dict()
def uniquePaths(self, m: int, n: int) -> int:
if (m, n) in self.visited:
return self.visited[(m, n)]
if m == 1 or n == 1:
return 1
cnt = self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
self.visited[(m, n)] = cnt
return cnt
- 记忆化递归
- 基本动态规划问题
- 空间复杂度可以进一步优化到O(n), 这会是一个考点
代码支持JavaScript,Python3
JavaScript Code:
/*
* @lc app=leetcode id=62 lang=javascript
*
* [62] Unique Paths
*
* https://leetcode.com/problems/unique-paths/description/
*/
/**
* @param {number} m
* @param {number} n
* @return {number}
*/
var uniquePaths = function(m, n) {
const dp = Array(n).fill(1);
for(let i = 1; i < m; i++) {
for(let j = 1; j < n; j++) {
dp[j] = dp[j] + dp[j - 1];
}
}
return dp[n - 1];
};
Python3 Code:
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
dp = [1] * n
for _ in range(1, m):
for j in range(1, n):
dp[j] += dp[j - 1]
return dp[n - 1]
复杂度分析
- 时间复杂度:$O(M * N)$
- 空间复杂度:$O(N)$
你可以做到比$O(M * N)$更快,比$O(N)$更省内存的算法么?这里有一份资料可供参考。
提示: 考虑数学