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。 

 你可以认为每种硬币的数量是无限的。 

 示例 1: 

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

 示例 2: 


输入:coins = [2], amount = 3
输出:-1 

 示例 3: 


输入:coins = [1], amount = 0
输出:0

 示例 4: 

输入:coins = [1], amount = 1
输出:1

 示例 5: 

输入:coins = [1], amount = 2
输出:2

 提示: 

 1 <= coins.length <= 12 
 1 <= coins[i] <= 231 - 1 
 0 <= amount <= 104 
public int coinChange(int[] coins, int amount) {
        int[] state = new int[amount + 1];

        for (int i = 1; i < state.length; i++) {
            int staMin = amount+1;
            for (int j = 0; j < coins.length; j++) {
                if (i - coins[j] >= 0 && state[i - coins[j]] != -1) {
                    staMin = Math.min(staMin, state[i - coins[j]]);
                }
            }
            state[i] = staMin + 1;
        }

        return state[amount] > amount ? -1 : state[amount];
    }

LeetCode 322 Coin Change找零钱——面试题_java基础