int change(int amount, vector<int>& coins) { vector<int> dp(amount + 1); dp[0] = 1; for (auto& coin : coins) { for (int i = coin; i <= amount; i++) { dp[i] += dp[i - coin]; } } return dp[amount]; } 分类: LeetCode 标签: 动态规划 好文要顶 关注我 收藏该...
https://leetcode.cn/problems/coin-change-2leetcode.cn/problems/coin-change-2 之前做过一个题是零钱兑换,给定不同面额的硬币,找出最少需要多少个硬币,使得其面值等于amount。很容易找出dp状态转移方程为dp[i] = min{dp[i-coin_1], dp[i-coin_2], ... ,dp[i-coin_n]} + 1。本题跟上一题...
classSolution(object):defchange(self, amount, coins):""" :type amount: int :type coins: List[int] :rtype: int """dp = [0] * (amount +1) dp[0] =1forcoinincoins:foriinrange(1, amount +1):ifcoin <= i: dp[i] += dp[i - coin]returndp[amount]classSolution(object):defchan...
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【LeetCode】518. Coin Change 2 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/coin-change-2/description/ 题目描述: You are given coins of different denominations and a total am...
322--Coin Change比较清晰的动态规划,状态转移方程和起始状态都是比较好找到的,但需要一系列的学习才能对这类动态规划问题熟悉。我会继续上传这个问题的变式的解法。, 视频播放量 49、弹幕量 0、点赞数 2、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 Nemesiscs, 作
这道题是之前那道Coin Change的拓展,那道题问我们最少能用多少个硬币组成给定的钱数,而这道题问的是组成给定钱数总共有多少种不同的方法。那么我们还是要使用DP来做,首先我们来考虑最简单的情况,如果只有一个硬币的话,那么给定钱数的组成方式就最多有1种,就看此钱数能否整除该硬币值。那么当有两个硬币的话...
【Leetcode】Coin Change 题目链接:https://leetcode.com/problems/coin-change/题目: -1.Example 1: coins =[1, 2, 5], amount =11 return3Example 2: coins =[2], amount =3 return-1. Note: You may assume that you have an infinite number of each kind of coin....
Coin Change Problem 1 & 2. What is a coin change problem? There are two types in this, almost similar:- 1.) Minimum number of coins- Coin Change 1 on Leetcode 2.) Maximum number of ways- Coin Change 2 on Leetcode So, we have been given acoinsarray which consists of different d...
https://leetcode.cn/problems/coin-changeleetcode.cn/problems/coin-change dp解法: 定义dp矩阵:dp[j]为凑成总金额为j时最少的硬币个数,则dp[amount]为我们最终要求的结果 定义dp转移方程:假设coins有coin_1、coin_2、...、coin_n,则dp[i] = min{dp[i-coin_1], dp[i-coin_2], ... ,dp[...