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55 lines
1.6 KiB
Python
55 lines
1.6 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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#######################################################################################
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# Leetcode 322 零钱兑换
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#
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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。
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#
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# 示例 1:
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# 输入: coins = [1, 2, 5], amount = 11
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# 输出: 3
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# 解释: 11 = 5 + 5 + 1
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#
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# 示例 2:
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# 输入: coins = [2], amount = 3
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# 输出: -1
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#
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# PS: PPT中关于凑零钱的问题和这题Leetcode类似，且本题比PPT上更通用
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#######################################################################################
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class Solution:
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def coinChange(self, coins, amount):
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"""
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:type coins: List[int],
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:type amount: int
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:rtype int
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(knowledge)
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思路：
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1. 使用动态规划；
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2. 定义f(n)为状态转移方程，表示凑n元所需的最少硬币数：
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-1 n < 0
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0 n == 0
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min{f(n - c) + 1 | c ∈ coins}
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"""
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dp = [amount + 1 for i in range(amount + 1)]
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dp[0] = 0
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for i in range(len(dp)):
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for coin in coins:
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if i - coin < 0:
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continue
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dp[i] = min(dp[i], dp[i - coin] + 1)
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return dp[amount] if dp[amount] != amount + 1 else -1
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if __name__ == '__main__':
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solution = Solution()
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print(solution.coinChange([1, 2, 5], 11), "= 3")
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print(solution.coinChange([2], 3), "= -1")
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