172 lines
4.0 KiB
Markdown
172 lines
4.0 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/unique-paths/description/
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## 题目描述
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```
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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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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).
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How many possible unique paths are there?
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```
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```
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Above is a 7 x 3 grid. How many possible unique paths are there?
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Note: m and n will be at most 100.
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Example 1:
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Input: m = 3, n = 2
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Output: 3
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Explanation:
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From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
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1. Right -> Right -> Down
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2. Right -> Down -> Right
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3. Down -> Right -> Right
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Example 2:
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Input: m = 7, n = 3
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Output: 28
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```
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## 思路
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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。
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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，
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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。
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代码大概是：
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JS Code:
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```js
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const dp = [];
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for (let i = 0; i < m + 1; i++) {
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dp[i] = [];
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dp[i][0] = 0;
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}
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for (let i = 0; i < n + 1; i++) {
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dp[0][i] = 0;
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}
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for (let i = 1; i < m + 1; i++) {
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for(let j = 1; j < n + 1; j++) {
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dp[i][j] = j === 1 ? 1 : dp[i - 1][j] + dp[i][j - 1]; // 转移方程
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}
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}
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return dp[m][n];
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```
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Python Code:
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```python
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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d = [[1] * n for _ in range(m)]
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for col in range(1, m):
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for row in range(1, n):
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d[col][row] = d[col - 1][row] + d[col][row - 1]
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return d[m - 1][n - 1]
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```
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**复杂度分析**
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- 时间复杂度：$O(M * N)$
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- 空间复杂度：$O(M * N)$
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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).
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具体代码请查看代码区。
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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：
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> 直接暴力递归的话会超时。
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Python3 Code:
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```python
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class Solution:
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visited = dict()
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def uniquePaths(self, m: int, n: int) -> int:
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if (m, n) in self.visited:
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return self.visited[(m, n)]
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if m == 1 or n == 1:
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return 1
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cnt = self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
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self.visited[(m, n)] = cnt
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return cnt
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```
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## 关键点
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- 记忆化递归
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- 基本动态规划问题
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- 空间复杂度可以进一步优化到O(n), 这会是一个考点
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## 代码
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代码支持JavaScript，Python3
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JavaScript Code:
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```js
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/*
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* @lc app=leetcode id=62 lang=javascript
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*
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* [62] Unique Paths
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*
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* https://leetcode.com/problems/unique-paths/description/
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*/
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/**
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* @param {number} m
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* @param {number} n
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* @return {number}
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*/
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var uniquePaths = function(m, n) {
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const dp = Array(n).fill(1);
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for(let i = 1; i < m; i++) {
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for(let j = 1; j < n; j++) {
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dp[j] = dp[j] + dp[j - 1];
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}
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}
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return dp[n - 1];
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};
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```
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Python3 Code:
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```python
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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dp = [1] * n
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for _ in range(1, m):
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for j in range(1, n):
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dp[j] += dp[j - 1]
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return dp[n - 1]
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```
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**复杂度分析**
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- 时间复杂度：$O(M * N)$
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- 空间复杂度：$O(N)$
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## 扩展
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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](https://leetcode.com/articles/unique-paths/)可供参考。
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> 提示： 考虑数学
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