322.CoinChange

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Note:

You may assume that you have an infinite number of each kind of coin.

ans[i]表示凑齐钱数i时需要的最小零钱数目

和题目494类似，

public int coinChange(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, amount + 1); // 初始化dp数组，初始值设为amount+1，保证后续更新时能正确找到最小值
    dp[0] = 0; // 金额为0时不需要硬币

    for (int i = 0; i < coins.length; i++) {
        for (int j = 1; j <= amount; j++) {
            if (coins[i] <= j) { 
                dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1); // 更新最少硬币数量
            }
        }
    }
    return dp[amount] == amount + 1 ? -1 : dp[amount]; // 如果无法凑成金额，则返回-1，否则返回结果
}