daily leetcode - unique-paths - ! - LouisLan的学习简记

2020-02-10 / 2020-02-10 / daily-leetcode

daily leetcode - unique-paths - !

题目地址

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

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?

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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

这道题让求所有不同的路径的个数，一开始还真把博主难住了，因为之前好像没有遇到过这类的问题，所以感觉好像有种无从下手的感觉。在网上找攻略之后才恍然大悟，原来这跟之前那道 Climbing Stairs 很类似，那道题是说可以每次能爬一格或两格，问到达顶部的所有不同爬法的个数。而这道题是每次可以向下走或者向右走，求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样，需要用动态规划 Dynamic Programming 来解，可以维护一个二维数组 dp，其中 dp[i][j] 表示到当前位置不同的走法的个数，然后可以得到状态转移方程为： dp[i][j] = dp[i - 1][j] + dp[i][j - 1]，这里为了节省空间，使用一维数组 dp，一行一行的刷新也可以，代码如下：

解法一：

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> dp(n, 1);
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[j] += dp[j - 1]; 
            }
        }
        return dp[n - 1];
    }
};

这道题其实还有另一种很数学的解法，参见网友 Code Ganker 的博客，实际相当于机器人总共走了 m + n - 2 步，其中 m - 1 步向右走，n - 1 步向下走，那么总共不同的方法个数就相当于在步数里面 m - 1 和 n - 1 中较小的那个数的取法，实际上是一道组合数的问题，写出代码如下：

解法二：

class Solution {
public:
    int uniquePaths(int m, int n) {
        double num = 1, denom = 1;
        int small = m > n ? n : m;
        for (int i = 1; i <= small - 1; ++i) {
            num *= m + n - 1 - i;
            denom *= i;
        }
        return (int)(num / denom);
    }
};

思路 2

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。

62.unique-paths-2

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).

mark

具体代码请查看代码区。

关键点解析

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];
};