【VIP】322. Coin Change 【M】【53】【动态规划】

最新推荐文章于 2021-09-08 18:44:44 发布

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最新推荐文章于 2021-09-08 18:44:44 发布

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分类专栏： leetcode Python 算法 DP 动态规划文章标签： leetcode python 动态规划

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解决硬币兑换问题，计算最少需要多少硬币组成指定金额。包括边界条件处理和无限硬币供应的假设。

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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:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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这个就是，从1开始，一直到amount，每个点都算一遍

然后对每个点 j ，计算coins数组的i（i<当前值），计算最小的 res[i-j] + 1 就是当前点的最佳方案

计算量还是挺大的

class Solution(object):
    def coinChange(self, coins, amount):
        if amount == 0:
            return 0

        res = [-1] * (max(max(coins),amount) + 1)

        for i in xrange(1,amount+1):
            if i  not in coins:
                minn = 99999999
                for j in coins:
                    if (i-j) < amount and res[i-j] > -1:
                        #print i-j
                        #minn = min(res[i-j]+1,minn) #为什么用这句会超时呢！？
                        if res[i-j] + 1< minn:
                            minn = res[i-j] + 1
                if minn != 99999999:
                    res[i] = minn
            else:
                res[i] = 1
        return res[amount]