322. Coin ChangeMedium Topics Companies You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made...
图片来自于https://leetcode-cn.com/problems/coin-change/solution/wan-quan-bei-bao-wen-ti-shou-hua-dp-table-by-shixu/ 值得一提的是, 以原问题 amount = 11, coins = [2,5,1] 为例子。 i=1代表的是只可以放 {coin = 2,count =1}下,可以放n个小于11的情况 ...
https://leetcode-cn.com/problems/coin-change 解题思路 动态规划,自底向上,太简单,不解释。 C++代码 classSolution{public:intcoinChange(vector<int>& coins,intamount){vector<int>dp(amount +1, amount +1); dp[0] =0;for(inti =1; i <= amount; i++) {for(autov : coins) {if(i >= v)...
依次向前推,直到r等于0或者小于0. public int coinChange(int[] coins, int amount) { if (amount < 0) return 0; return coinChangeCore(coins, amount, new int[amount]); } private int coinChangeCore(int[] coins, int amount, int[] count) { if (amount < 0) return -1; if (amount ==...
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. ...
Leetcode 322. Coin Change 硬币找零问题 MaRin 菜鸡一只 给你一个整数数组 coins ,表示不同面额的硬币;以及一个整数 amount ,表示总金额。计算并返回可以凑成总金额所需的 最少的硬币个数 。如果没有任何一种硬币组合能组成总金额,返回 -1 。你可以认为每种硬币的数量是无限的。对应leetcode链接为: https:/...
LeetCode 322. Coin Change 简介:给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额,返回 -1。 Description \You are given coins of different denominations and a total amount of money amount. Write a ...
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
Coin Change【硬币找零】 技术标签: leetcode一、题目 英文:Coin Change 中文:硬币找零 二、内容要求 英文:You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to ma......
LeetCode 322. Coin Change 题目 动态规划 class Solution { public: int dp[10005]; int coinChange(vector<int>& coins, int amount) { memset(dp,-1,sizeof(dp)); dp[0] = 0; for(int i=1;i<=amount;i++) { for(int j=0;j<coins.size();j++)...