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:
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];
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。
关键点
- 空间复杂度可以进一步优化到O(n), 这会是一个考点
- 基本动态规划问题
代码
/* * @lc app=leetcode id=62 lang=javascript * * [62] Unique Paths * * https://leetcode.com/problems/unique-paths/description/ * * algorithms * Medium (46.53%) * Total Accepted: 277K * Total Submissions: 587.7K * Testcase Example: '3\n2' * * 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 * * START */ /** * @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]; };LeetCode题解计算机为什么是基于二进制的?
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