Coin Change

最新推荐文章于 2020-05-18 23:50:55 发布

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最新推荐文章于 2020-05-18 23:50:55 发布

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原文链接：http://www.cnblogs.com/liudebo/p/9276800.html

本文介绍了一种使用动态规划解决硬币找零问题的方法，即如何用最少数量的硬币组合达到指定金额。通过具体例子展示了算法的实现过程，并提供了一个简洁高效的 Java 代码实现。

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:

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

Example 2:

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

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

Solution:

Using dynamic programming, a[i] for the fewest coins to meet i.

class Solution {
    public int coinChange(int[] coins, int amount) {
        int max = amount+1;
        int[] a = new int[max];
        Arrays.fill(a,1,max,max);
        for (int i = 1; i < max; i++) {
            for (int j = 0; j < coins.length; j++) {
                if (i>=coins[j] && a[i-coins[j]] + 1 < a[i]) {
                    a[i] = a[i-coins[j]] + 1;
                }
            }
        }
        return a[amount] == max ? -1 : a[amount];
    }
}