518. Coin Change 2

最新推荐文章于 2021-02-07 11:55:26 发布

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本文介绍了一个计算特定金额下不同硬币组合数量的问题。通过动态规划的方法，使用一维数组记录每种金额的组合数，并逐步更新数组以得到最终答案。

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You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000
1 <= coin <= 5000
the number of coins is less than 500
the answer is guaranteed to fit into signed 32-bit integer
Example 1:

Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
Example 3:

Input: amount = 10, coins = [10]
Output: 1

代码解析：动态规划；构造一个一维数组dp[amount+1]用来记录小于等于amount的每一个整数的构成种类数.逐一遍历coin[n]中的数c=coin[j],dp[i] += dp[i-c]：意思就是如果c存在于coin[n]中那么构成i的种类就等于构成i-c的种类加上原来构成i的种类


public class Solution {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
int[] coins = {1,2,5};
System.out.println(change(5, coins));
    }
    public static  int change(int amount, int[] coins) {
        int[] dp = new int[amount+1];
        dp[0] = 1;
        for(int j = 0;j<coins.length;j++){
            int c = coins[j];
            for(int i = c;i<amount+1;i++){
                dp[i] += dp[i-c];
            }
        }

        return dp[amount];

    }
}