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] ...
classSolution(object):def__init__(self):self.ans=float('inf')defcoinChange(self,coins,amount):""":type coins: List[int]:type amount: int:rtype: int"""ifnotcoins:return-1ifnotamount:return0coins.sort(reverse=True)defcoin_change(coins,s,amount,count):"""深度优先搜索树:param coins:硬...
代码如下: 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(objec...
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, return-1. Example 1: coins =[1, ...
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 ...
简介:Leetcode-Medium 322. Coin Change 题目描述 假设给你不同面额的硬币和一个金额amount。编写一个函数来计算构成该金额amount所需的最少数量的硬币。如果这笔钱不能由任何硬币组合成,则返回-1。 思路 动态规划: 假设amount为10,硬币面额为[1,2,5,10],用dp[i]来表示金额i所需要的最少硬币数,那么显然dp...
Coin Change dp[i]=min(dp[i],dp[i-coin]+1) 另外我们需要初始化dp,假设每一个硬币的面额都大于amount,此时我们是找不出硬币组合的,那么dp[amount]=-1,显然我们不能初始化所有值为...Change - LeetCode Discuss Leetcode No.322( Coin Change) 心得(Medium) – ChingYuanYang – Medium DP: 322......
Python meysam81/change-coin Sponsor Star0 coin dispenser machine, returning the optimal result in O(numberOfCoins * targetValue) javaalgorithmcoin-changecoin-changercoin-dispenser UpdatedJun 6, 2018 Java All solutions to the problems in "Dynamic Programming I", the LeetCode dynamic programming study...
Python3 实现: 代码语言:javascript 复制 classSolution:defchange(self,amount:int,coins:List[int])->int:dp=[1]+[0]*amountforcoinincoins:forjinrange(coin,amount+1):dp[j]+=dp[j-coin]# dp[j]是所有 coins 累加的结果returndp[-1]print(Solution().change([2,1],3))#2(111,12) ...
Python3 实现: class Solution: # 方法1:DP, 完全背包 def coinChange(self, coins: List[int], amount: int) -> int: dp = [0] + [float("inf")] * amount for coin in coins: for j in range(coin, amount+1): # 从 coin 开始递增到 amount ...