凑零钱问题

题⽬ ：
给你 k 种⾯值的硬币， ⾯值分别为 c1, c2 … ck ， 每种硬
币的数量⽆限， 再给⼀个总⾦额 amount ， 问你最少需要⼏枚硬币凑出这个
⾦额， 如果不可能凑出， 算法返回 -1。

# -*- coding: utf-8 -*-
"""
Created on Wed Mar  3 14:14:19 2021

@author: dujidan
"""

from collections import defaultdict


# 递归
def coinChange(coins, amount):
    def dp(n):
        #base case
        if n == 0:
            return 0
        if n < 0:
            return -1
        #求最小值， 所以初始化为正无穷
        res = float('inf')
        for coin in coins:
            subproblem = dp(n - coin)
            if subproblem == -1:
                continue
            res = min([res, 1 + subproblem])
        return res if res != float('inf') else -1
    return dp(amount)


# 递归 备忘录
def coinChange(coins, amount):
    meno = {}

    def dp(n):
        # 备忘录 避免重复计算
        if n in meno:
            return meno[n]
        # base case
        if n == 0:
            return 0
        if n < 0:
            return -1
        # 求最小值， 所以初始化为正无穷
        res = float('inf')
        for coin in coins:
            subproblem = dp(n - coin)
            if subproblem == -1:
                continue
            res = min([res, 1 + subproblem])
        meno[n] = res if res != float('inf') else -1
        return meno[n]
    return dp(amount)


# 数组的迭代解法
def coinChange(coins, amount):
    dp = defaultdict(int)
    dp[0] = 0
    for i in range(1, amount + 1):
        dp[i] = amount + 1
        for coin in coins:
            if i - coin < 0:
                continue
            dp[i] = min([dp[i], 1 + dp[ i - coin]])
    return ( -1 if dp[amount] == -1 else dp[amount])


coins = [1, 3, 5]
amount = 11
print(coinChange(coins, amount))

参考：labuladong 公众号