[LeetCode] Find minimum number of coins (Dynamic Programming)

最新推荐文章于 2021-03-12 15:35:30 发布

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本文探讨了硬币找零问题的两种算法解决方案：递归和动态规划。通过实例展示了如何寻找最少数量的硬币来组成特定金额。递归方法直接但效率低下，而动态规划方法则提供了一种更高效解决问题的途径。

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Ex1:

Input: coins[] = {25, 10, 5}, V = 30
Output: minimum 2 coins required

Ex2:

Input: coins[] = {9, 6, 5, 1}, V = 11
Output: minimum 2 coins required

Recursive Solution

If V == 0, then 0 coins required.
If V > 0
   minCoin(coins[0..m-1], V) = min {1 + minCoins(V-coin[i])} 
                               where i varies from 0 to m-1 
                               and coin[i] <= V

Python Naive Recursive Solution

class Solution:
    def coinChange(self, coins, amount):
        if amount == 0:
            return 0
        for i in range(0, len(coins)):
            mincoins = min(1+self.coinChange(amount - coins[i]))
            if mincoins + 1 < mincoins:
                res = mincoins + 1
        return res

DP solution

def coinChangeDP(self, coins, amount):
    MAX = float('inf')
    dp = [0] + [MAX]*amount

    for i in range(1, amount + 1):
        dp[i] = min([dp[i - c] if i - c >= 0 else MAX for c in coins]) + 1

    return [dp[amount], -1][dp[amount] == MAX]