Day 17 LeetCode 322. Coin Change

换个口味来写 LeetCode，并且挑战一下不太擅长的动态规划。

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return 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.

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

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

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

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

class Solution:

    def solve(self, coins: List[int], amount: int):

        box = [amount+1 for i in range((amount + 1))]
        box[0] = 0
        for i in range(1, amount + 1):
            for coin in coins:

                if coin <= i:
                    box[i] = min(box[i], box[i - coin] + 1)


        return -1 if box[amount] > amount else box[amount]
        pass

    def coinChange(self, coins: List[int], amount: int) -> int:

        if amount == 0:
            return 0

        return self.solve(coins, amount)

在这一题中最重要的就是 11 行的 box[i - coin] + 1，他代表继承之前的硬币数。

比如说现在要找总合 i == 10 的，当 coin == 3 时，
会找 box[7] 的硬币数并+1，
因为任何一组能构成总和 7 的硬币，只要加这一枚硬币 3 总和就是 10。