代码(Python3) class Solution: def coinChange(self, coins: List[int], amount: int) -> int: # dp[i] 表示凑出 i 所需的最少硬币数量, # 初始化为 amount + 1 ,表示当前还凑不出 dp: List[int] = [amount + 1] * (amount + 1) # 最开始只能确认不需要任何硬币就可以凑出 0 dp[0] ...
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 ...
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
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)...
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【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....
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...
[LeetCode] 322. Coin Change 硬币找零 You are given coins of different denominations and a total amount of moneyamount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins...
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++)...