Skip to content

Latest commit

 

History

History
171 lines (124 loc) · 3.99 KB

62.unique-paths.md

File metadata and controls

171 lines (124 loc) · 3.99 KB
Raw

题目地址

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

题目描述


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

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

关键点

  • 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

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：$O(N)$

扩展

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