[leetcode] 322. Coin Change @ python

最新推荐文章于 2022-11-14 20:33:38 发布

原创最新推荐文章于 2022-11-14 20:33:38 发布 · 286 阅读

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本文探讨了硬币找零问题，即如何使用最少数量的硬币组合来达到特定金额。通过动态规划的方法，我们构建了一个dp数组，用于记录到达每个金额所需的最少硬币数。文章详细解释了状态转移方程，并提供了Python实现代码。

原题

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.

解法

参考: https://www.youtube.com/watch?v=uUETHdijzkA
动态规划, 构造dp数组, 长度为amount+1, dp[i]表示构成面额为i 时最少需要的硬币数量, 长度为amount+1是由于我们需要构成面额amount. 初始化dp[0] = 0, 遍历coins, 更新dp, 遍历完成后, 如果dp[-1]不为无穷大, 则返回dp[-1].
状态转移方程:
dp[i]表示构成面额为i 时最少需要的硬币数量 = 构成面额为i -coin最少需要的硬币数量 + coin这一枚硬币.

dp[i] = min(dp[i], dp[i-coin] + 1)

代码

class Solution(object):
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        dp = [float('inf')]*(amount+1)
        dp[0] = 0
        for coin in coins:
            for i in range(coin, amount+1):
                if dp[i-coin] != float('inf'):
                    dp[i] = min(dp[i], dp[i-coin] + 1)
        return dp[-1] if dp[-1] != float('inf') else -1