[LeetCode - Dynamic Programming] Coin Change

硬币找零问题

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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文章标签：数据结构与算法

原文链接：https://segmentfault.com/a/1190000007504757

解决硬币找零问题，计算组成特定金额所需的最少硬币数量。采用动态规划方法，通过状态转移方程优化求解过程。

Coin Change
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.

1.解题思路
动态规划，用dp[i]表示总价为i的最小纸币张数，很容易想到状态转移方程:
dp[i]=Math.min(dp[i-coins[0]]+1,dp[i-coins[1]]+1,...,dp[i-coins[n-1]]+1),当然前提是i要大于纸币金额数。
dp[i-coins[0]]+1: 表示取一张coins[0]面额加上合计为i-coins[0]的最小纸币数。
另题目要求无法合计出的金额i，dp[i]要返回-1，所以要作特殊处理，否则就会返回元素初始化值0.

2.代码

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