518. Coin Change 2

最新推荐文章于 2022-02-08 12:35:02 发布

转载最新推荐文章于 2022-02-08 12:35:02 发布 · 299 阅读

本文介绍了一种解决组合找零问题的有效算法。该算法利用动态规划思想，通过一个示例详细展示了如何计算不同面额硬币组成特定金额的方法数量。此外，还提供了一种递归方法来解决相同的问题。

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

public class Solution {  
    public int change(int amount, int[] coins) {  
        if (coins == null || coins.length == 0) {  
            return amount == 0? 1: 0;  
        }  
        int[] dp = new int[amount + 1];  
        dp[0] = 1;  
        for (int i = 0; i < coins.length; i ++) {  
            for (int j = 1; j <= amount; j ++) {  
                if (j >= coins[i]) {  
                    dp[j] += dp[j - coins[i]];  
                }  
            }  
        }  
        return dp[amount];  
    }  
}

递归：

 public int change(int amount, int[] coins) {
        if (amount < 0 || coins == null ) {
            return 0;
        }
        return getNum(coins, coins.length - 1, amount);
    }

    public int getNum(int[] coins, int i, int remain) {
        if (remain <= 0 || i < 0) {
            if (remain == 0) {
                return 1;
            }
            return 0;
        }
        return getNum(coins, i - 1, remain) + getNum(coins, i, remain - coins[i]);
    }