#322. 零钱兑换

给定不同面额的硬币coins和一个总金额amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回-1。

示例1:

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输入: coins = [1, 2, 5], amount = 11
输出: 3
解释: 11 = 5 + 5 + 1

示例2:

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输入: coins = [2], amount = 3
输出: -1

说明:
你可以认为每种硬币的数量是无限的。

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package xyz.onns.leetcode;

import java.util.Arrays;

/**
 * @Author Onns
 * @Date 2020/9/15 12:06 PM
 * @Version 1.0
 * @Website https://onns.xyz/blog/
 */
public class CoinChange {
	public static void main(String[] args) {
		new CoinChange().new Solution().test();
	}

	class Solution {
		public int coinChange(int[] coins, int amount) {

			int n = coins.length;
			if (n == 0) return -1;
			int[] dp = new int[amount + 1];
			Arrays.fill(dp, amount + 1);

			// base case:
			// dp[0] = 0
			dp[0] = 0;

			// all conditions:
			// amount from 1 to amount
			// coin from 0 to coins.length - 1
			for (int i = 1; i <= amount; i++) {
				for (int j = 0; j < n; j++) {
					// state transition:
					// f(i) = f(i-coins[i]) + 1, if f(i-coins[i]) + 1 < f(i)
					// f(i) = f(i), otherwise
					int t = i - coins[j];
					if (t >= 0) {
						dp[i] = Math.min(dp[t] + 1, dp[i]);
					}
				}
			}
			return dp[amount] == (amount + 1) ? -1 : dp[amount];
		}

		void test() {
			System.out.println(coinChange(new int[]{1, 2, 5}, 5));
			System.out.println(coinChange(new int[]{1, 2, 5}, 10));
			System.out.println(coinChange(new int[]{1, 2, 5}, 11));
			System.out.println(coinChange(new int[]{2}, 3));
		}
	}
}