【LeetCode】【322. Coin Change】(python版)

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本文链接： https://blog.csdn.net/qq_20141867/article/details/81264178

Description：
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 make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:

Input: coins = [1, 2, 5], amount = 11
Output: 3 
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Note:
You may assume that you have an infinite number of each kind of coin.

思路：
本题意思是有一堆不同面额的硬币，问最少取多少枚硬币，可以凑够想要的面值，每种硬币数无限。

假设dp[i]表示凑够i元所需要的最少硬币数，一共有n种面值硬币，那么$dp[i] = min(dp[i-coins[0]],dp[i-coins[1]],...dp[i-coins[k])+1，其中coins[k]<=i$

比如例1中

dp[0] = 0
dp[1] = 1
dp[2] = min{dp[2-1]}+1
dp[3] = min{dp[3-1],dp[3-2]}+1
dp[4] = min{dp[4-1],dp[4-2]}+1
...
dp[11] = min{11-1},dp[11-2],dp[11-5]}+1

实现代码如下：

class Solution(object):
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        n = len(coins)
        # dp[i]表示amount=i需要的最少coin数
        dp = [float("inf")] * (amount+1)
        dp[0] = 0
        for i in range(amount+1):
            for j in range(n):
                # 只有当硬币面额不大于要求面额数时，才能取该硬币
                if coins[j] <= i:
                    dp[i] = min(dp[i], dp[i-coins[j]]+1)
        # 硬币数不会超过要求总面额数，如果超过，说明没有方案可凑到目标值
        return dp[amount] if dp[amount] <= amount else -1