【LeetCode】518. Coin Change 2 解题报告（Python）

【LeetCode】518. Coin Change 2 解题报告（Python）

id： fuxuemingzhu

个人博客： ​​​http://fuxuemingzhu.cn/​​​

题目地址：​​https://leetcode.com/problems/coin-change-2/description/​​

题目描述：

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

1. 1. 
      
2. 0 <= amount <= 5000
3. 1 <= coin <= 5000
4. the number of coins is less than 500
5. the answer is guaranteed to fit into signed 32-bit integer
   1. 
      

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

题目大意

有一堆一定面额的硬币，问有多少种可以组成amount的方案。假定硬币的数量是不限量的。

解题方法

DP。第一感觉是完全背包问题，但其实由于没有重量和价值的对应关系，所以不一样。

生成了一个一维数组dp，dp[i]代表了生成总价值为i有多少方案。

对已有的所有面值的硬币进行遍历，其实思路很简单：dp[i] += dp[i - coin]，价值为i的解决方案应该加上价值为i - coin的解决方案。

时间复杂度是O(L * A)，空间复杂度是O(A); A = amount.

代码如下：

class Solution(object):
    def change(self, amount, coins):
        """
        :type amount: int
        :type coins: List[int]
        :rtype: int
        """
        dp = [0] * (amount + 1)
        dp[0] = 1
        for coin in coins:
            for i in range(1, amount + 1):
                if coin <= i:
                    dp[i] += dp[i - coin]
        return dp[amount]

参考资料：

​​https://www.youtube.com/watch?v=jaNZ83Q3QGc​​

日期

2018 年 9 月 25 日 —— 美好的一周又开始了，划重点，今天是周二