Question

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.

Code

public class Solution {
    public int coinChange(int[] coins, int amount) {
        if (coins == null || coins.length == 0 || amount == 0) return 0;
        
        int[] dp = new int[amount + 1];
        
        for (int i = 1; i < dp.length; i++) {
            dp[i] = Integer.MAX_VALUE;
            for (int j = 0; j < coins.length; j++) {
                if (i >= coins[j] && dp[i - coins[j]] != Integer.MAX_VALUE) {
                    dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        
        return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
    }
}

Solution

動態規劃。

用dp存儲硬幣數量，dp[i] 表示湊齊錢數 i 需要的最少硬幣數，那么湊齊錢數 amount 最少硬幣數為：固定錢數為 coins[j] 一枚硬幣，另外的錢數為 amount - coins[j] 它的數量為dp[amount - coins[j]]，j 從0遍歷到coins.length - 1