_322_CoinChange.java

package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _322_CoinChange {
    /**
     * 322. Coin Change
     * ou 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.

     dp[] amount 需要多少coins

     min = Math.min(min, dp[i - coins[j]] + 1);

     * time : O(n*amount)
     * space : O(amount)
     * @param coins
     * @param amount
     * @return
     */
    public int coinChange(int[] coins, int amount) {
        if (amount == 0) return 0;
        if (coins == null || coins.length == 0) return -1;

        int[] dp = new int[amount + 1];
        for (int i = 1; i <= amount; i++) {
            int min = Integer.MAX_VALUE;
            for (int j = 0; j < coins.length; j++) {
                if (i >= coins[j] && dp[i - coins[j]] != -1) {
                    min = Math.min(min, dp[i - coins[j]] + 1);
                }
            }
            dp[i] = min == Integer.MAX_VALUE ? -1 : min;
        }
        return dp[amount];
    }
}