硬币更改动态编程问题自上而下的记忆方法

Question

我目前正在研究有关leetcode的硬币找零动态编程问题-https://leetcode.com/problems/coin-change/。

这是问题陈述：

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

[我试图实现一种自上而下的记忆方法，在该方法中，我保留一个长度数量的数组，其中每个索引代表可以用来制作该数量的最小硬币数量。

这是我的Java代码：

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];

        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        int min = coinChange(coins, amount, dp);

        return min == Integer.MAX_VALUE ? -1 : min;
    }

    private int coinChange(int[] coins, int amount, int[] dp) {
        if (amount < 0) {
            return Integer.MAX_VALUE;
        }
        if (amount == 0) {
            return 0;
        }

        if (dp[amount] != Integer.MAX_VALUE) {
            return dp[amount];
        }

        int min = Integer.MAX_VALUE;
        for (int i = 0; i < coins.length; i++) {
            int val = coinChange(coins, amount - coins[i], dp);

            if (val != Integer.MAX_VALUE) {
                min = Math.min(min, val + 1);
            }
        }

        dp[amount] = min;
        return min;
    }
}

我认为这是针对此问题进行动态编程的正确方法，但是我在leetcode上超过了时限。

动态编程是错误的方法吗？如果是这样，请您解释哪里出了问题？

非常感谢。

Answer 1

这是我的解决方案版本。这也很容易理解！

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];

        Arrays.fill(dp, 0);

        int min = coinChange(coins, amount, dp);
        return min;
    }

    private int coinChange(int[] coins, int amount, int[] dp) {
        if (amount < 0) {
            return -1;
        }
        if (amount == 0) {
            return 0;
        }

        if (dp[amount]!=0) {
            return dp[amount];
        }

        int minimum = Integer.MAX_VALUE;
        for (int i = 0; i < coins.length; i++) {
            int val = coinChange(coins, amount - coins[i], dp);

            if (val >= 0 && val < minimum) {
                minimum = val + 1;
            }
        }
        dp[amount] = (minimum == Integer.MAX_VALUE) ? -1 : minimum;
        return dp[amount];
    }
}