LeetCode—322. Coin Change

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本文介绍了一个经典的CoinChange问题，探讨了如何使用动态规划方法来计算组成特定金额所需的最少硬币数量。通过两个示例说明了算法的工作原理，并提供了一段Java代码实现。

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322. Coin Change

Total Accepted: 837 Total Submissions: 3093 Difficulty: Medium

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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思路：动态规划可以轻松解决。注意题目没有说sort是有序的。

GitHub地址：https://github.com/corpsepiges/leetcode

有Java版本和Python版本，求star。

public class Solution {
	int[] array;
	int minValue;
	Map<Integer,Integer> map=new HashMap<Integer, Integer>();
    public int coinChange(int[] coins, int amount) {
    	Arrays.sort(coins);
    	minValue=coins[0];
    	array=coins;
    	return f(amount);
    }
    public int f(int amount){
    	if (map.get(amount)!=null) {
			return map.get(amount);
		}
    	if (amount==0) {
			return 0;
		}
    	if (amount<minValue) {
			return -1;
		}
    	int min=Integer.MAX_VALUE;
    	for (int i = 0; i < array.length; i++) {
			int v=f(amount-array[i]);
			if (v!=-1&&v<min) {
				min=v;
			}
		}
    	int ans=0;
    	if (min==Integer.MAX_VALUE) {
			ans=-1;
		}else{
			ans=1+min;
		}
    	map.put(amount,ans);
    	return ans;
    }
}