669. 换硬币

给出不同面额的硬币以及一个总金额. 写一个方法来计算给出的总金额可以换取的最少的硬币数量. 如果已有硬币的任意组合均无法与总金额面额相等, 那么返回 -1.

样例

给出 coins = [1, 2, 5], amount = 11
返回 3 (11 = 5 + 5 + 1)
给出 coins = [2], amount = 3
返回 -1

注意事项

你可以假设每种硬币均有无数个

class Solution:
    """
    @param coins: a list of integer
    @param amount: a total amount of money amount
    @return: the fewest number of coins that you need to make up
    """
    
    '''
    def coinChange(self, coins, amount):
        # write your code here
        if len(coins) == 0 or amount < 0:
            return -1
        
        n = len(coins)
        MAX = float('inf')
        dp = [[0] * (amount+1) for _ in range(n)]
        
        #初始化dp[0]
        for i in range(1, amount+1):
            dp[0][i] = MAX
            if i - coins[0] >=0 and dp[0][i-coins[0]] != MAX:
                dp[0][i] = dp[0][i-coins[0]] + 1
        
        left = 0
        for i in range(1, n):
            for j in range(1, amount+1):
                left = MAX
                if j - coins[i] >= 0 and dp[i][j-coins[i]] != MAX:
                    left = dp[i][j - coins[i]] + 1
                
                dp[i][j] = min(left, dp[i-1][j])
        
        return dp[n-1][amount] if dp[n-1][amount] != MAX else -1
    '''
    
    #空间压缩方法
    def coinChange(self, coins, amount):
        # write your code here
        if len(coins) == 0 or amount < 0:
            return -1
        
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0
        
        for i in range(1, amount+1):
            #遍历银币种类，大额满足条件时，会覆盖小额
            for coin in coins:
                if i >= coin and dp[i] > dp[i-coin] + 1:
                    dp[i] = dp[i-coin] + 1
        
        return dp[-1] if dp[-1] <= amount else -1