322. Coin Change

最新推荐文章于 2024-03-19 21:22:22 发布

原创最新推荐文章于 2024-03-19 21:22:22 发布 · 227 阅读

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本文介绍了一种解决硬币找零问题的方法，通过动态规划算法计算组成特定金额所需的最少硬币数量。如果无法使用给定面额的硬币组合成该金额，则返回-1。文中提供了一个具体实现示例。

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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.

递推方程 dp[i]=min{dp[i-j]}+1 1<j<=i

public static int coinChange(int[] coins, int amount)
	{
		int len=coins.length;
		if(len<1)
			return -1;
		
		long[] dp=new long[amount+1];
		for(int i=0;i<len;i++)
			{
				int num=coins[i];
				if(num<=amount)
					dp[num]=1;
			}
				
		for(int i=1;i<=amount;i++)
		{
			long min=Integer.MAX_VALUE;
			for(int j=0;j<len;j++)
				if(i-coins[j]>=0)
					min=Math.min(min, dp[i-coins[j]]+1);
			dp[i]=min;
		}
		
		long ret=dp[amount];
		return (int) (ret==Integer.MAX_VALUE?-1:ret);
	}