找零兑换问题

问题一 求组成目标金额的最小硬币数

解法一：递归

class Solution:
    def __init__(self):
        self.minn = float('inf')
    def coinChange(self, coins, amount):
        def Change(coins, num, rem):
            if rem == 0:
                self.minn = min(self.minn, num)
                return
            for coin in coins:
                if rem >= coin:
                    Change(coins, num+1, rem-coin)
        Change(coins, 0, amount)
        if self.minn == float('inf'):
            return -1
        return self.minn

解法二：动态规划

class Solution:
    def coinChange(self, coins, amount):
        if amount == 0:
            return 0
        dp = [float('inf') for _ in range(amount+1)]
        dp[0] = 0
        for i in range(len(coins)):
            for j in range(coins[i], amount+1):
                dp[j] = min(dp[j], dp[j-coins[i]]+1)
        if dp[amount] == float('inf'):
            return -1
        return dp[amount]

问题二 求组成目标金额的不同硬币数量

一、组合

硬币数组为外层循环。对于每一种硬币，遍历一遍dp数组。这样可以避免重复，即[1, 1, 2]和[1, 2, 1]的情况。

class Solution:
    def change(self, amount, coins):
        dp = [0 for _ in range(amount+1)]
        dp[0] = 1
        for coin in coins:
            for j in range(coin, amount+1):
                dp[j] += dp[j-coin]
        return dp[amount]

二、排列

将遍历dp数组作为外层循环，遍历每一种硬币作为内层循环。

class Solution:
    def change(self, amount, coins):
        dp = [0 for _ in range(amount+1)]
        dp[0] = 1
        for j in range(amount+1):
            for coin in coins:
                if j >= coin:
                    dp[j] += dp[j-coin]
        return dp[amount]