Python version-LeetCode Learning: 518. Change Exchange II

Given coins of different denominations and a total amount. Write a function to calculate the number of coin combinations that can make up the total amount. Suppose there are infinite coins of each denomination. 

Example 1:

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

Source: LeetCode
Link: https://leetcode-cn.com/problems/coin-change-2

Method 1: Dynamic programming

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        coin_len=len(coins)
        dp=[ [0 for _ in range(amount+1)] for _ in range(coin_len+1) ]

        for i in range(coin_len+1):
            dp[i][0]=1

        # 扫描全部coins和amount的组合
        for  i in range(1,coin_len+1):
            for j in range(1,amount+1):
                # 如果你不把这第 i 个物品装入背包，也就是说你不使用 coins[i] 这个面值的硬币，
                # 那么凑出面额 j 的方法数 dp[i][j] 应该等于 dp[i-1][j]，继承之前的结果。
                # 如果你把这第 i 个物品装入了背包，也就是说你使用 coins[i] 这个面值的硬币，那
                # 么 dp[i][j] 应该等于 dp[i][j-coins[i-1]]
                # 
                if j-coins[i-1]>=0:
                    dp[i][j]=dp[i-1][j]+dp[i][j-coins[i-1]]
                else:
                    dp[i][j]=dp[i-1][j]
        
        return dp[coin_len][amount]

Method 2: Shrink a two-dimensional array into a one-dimensional array

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        coin_len=len(coins)
        # 压缩两维数组为1维数组
        dp=[0 for _ in range(amount+1)]
        dp[0]=1

        for i in range(0,coin_len):
            for j in range(1,amount+1):
                if (j-coins[i-1])>=0:
                    dp[j]=dp[j]+dp[j-coins[i-1]]

        return dp[amount]

 Method 3:

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        coin_len=len(coins)
        
        # 特殊条件判断
        if amount == 0:
            return 1
        # 根据Python特性重新构建新的迭代关系，并修改原有的状态转移公式
        dp = [0] * (amount + 1)
        dp[0] = 1
        for coin in coins:
            for i in range(coin, amount + 1):
                dp[i] = dp[i] + dp[i - coin]
        return dp[amount]