leetcode 322. 零钱兑换

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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。

示例 1:

输入: coins = [1, 2, 5], amount = 11
输出: 3 
解释: 11 = 5 + 5 + 1

示例 2:

输入: coins = [2], amount = 3
输出: -1

说明:
你可以认为每种硬币的数量是无限的。

  • dp solution o(n^2)
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class Solution(object):
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        dp = [0]
        for money in range(1,amount+1):
            dp.append(float('inf'))
            for coin in coins:
                if money-coin >= 0:
                    dp[-1] = min(dp[-1],dp[money - coin]+1)
        if dp[-1] == float('inf'):return -1
        return dp[-1]
  • dfs solution o(n*amount/coins[i])
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class Solution(object):
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        if amount == 0: return 0
        coins.sort(reverse=True)
        self.best = float('inf')
        def dfs(start, coins, amount, cur):
            if start == len(coins)-1:
                if amount%coins[start] == 0:
                    self.best = min(self.best, cur + amount/coins[start])
            else:
                for i in range(amount/coins[start], -1, -1):
                    if cur + i >= self.best: break
                    dfs(start+1, coins, amount-i*coins[start], cur+i)
        dfs(0, coins, amount, 0)
        return self.best if self.best != float('inf') else -1