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(dòng)態(tài)規(guī)劃。

用dp存儲(chǔ)硬幣數(shù)量，dp[i] 表示湊齊錢(qián)數(shù) i 需要的最少硬幣數(shù)，那么湊齊錢(qián)數(shù) amount 最少硬幣數(shù)為：固定錢(qián)數(shù)為 coins[j] 一枚硬幣，另外的錢(qián)數(shù)為 amount - coins[j] 它的數(shù)量為dp[amount - coins[j]]，j 從0遍歷到coins.length - 1