[Swift]LeetCode518. 零钱兑换 II | Coin Change 2

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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. 

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 

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

---

给定不同面额的硬币和一个总金额。写出函数来计算可以凑成总金额的硬币组合数。假设每一种面额的硬币有无限个。  

示例 1:

输入: amount = 5, coins = [1, 2, 5]
输出: 4
解释: 有四种方式可以凑成总金额:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

示例 2:

输入: amount = 3, coins = [2]
输出: 0
解释: 只用面额2的硬币不能凑成总金额3。

示例 3:

输入: amount = 10, coins = [10] 
输出: 1 

注意:

你可以假设：

• 0 <= amount (总金额) <= 5000
• 1 <= coin (硬币面额) <= 5000
• 硬币种类不超过 500 种
• 结果符合 32 位符号整数

---

52ms

class Solution {
    func change(_ amount: Int, _ coins: [Int]) -> Int {
        var memo = Array(repeating: 0, count: amount + 1)
        memo[0] = 1
        
        var coins = coins.sorted()
        
        for coin in coins where coin <= amount {
            for higherAmount in coin...amount {
                memo[higherAmount] += memo[higherAmount - coin]
            }
        }
        
        return memo[amount]
    }
}

---

184ms

class Solution {
    var cache: [Int: Int] = [:]
    func change(_ amount: Int, _ coins: [Int]) -> Int {
        return helper(amount, 0, coins.sorted(by: >))
    }
    
    private func helper(_ amount: Int, _ index: Int, _ coins: [Int]) -> Int {
        if amount == 0 { 
            return 1 
        }
        if index >= coins.count { return 0 }
        if let some = cache[5000 * index + amount] {
            return some
        }
        
        var res = 0
        for i in index..<coins.count {
            let coin = coins[i]
            if coin <= amount {
                res += helper(amount - coin, i, coins)
            }
        }
        cache[5000 * index + amount] = res
        return res
    }
}

---

316ms

class Solution {
    func change(_ amount: Int, _ coins: [Int]) -> Int {
        var row = [1]
        row.append(contentsOf: [Int](repeating: 0, count: amount))
        var dp = [[Int]](repeating: row, count: coins.count + 1)
        // i: number of coins, the last index in this range is i - 1
        guard 1 <= coins.count else {
            return dp[coins.count][amount]
        }
        for i in 1...coins.count {
            guard 1 <= amount else {
                continue
            }
            for j in 1...amount {
                dp[i][j] = dp[i - 1][j] + (j < coins[i - 1] ? 0 : dp[i][j - coins[i - 1]])
            }
        }
        return dp[coins.count][amount]
    }
}

---

332ms

class Solution {
    func change(_ amount: Int, _ coins: [Int]) -> Int {
        let a = amount + 1
        let c = coins.count + 1
        
        var dp = [[Int]](repeating: [Int](repeating: 0, count: c), count: a)
        
        for j in 0..<c {
            dp[0][j] = 1
        }
        
        for i in 1..<a {
            for j in 1..<c {
                if coins[j - 1] <= i {
                    dp[i][j] = dp[i][j - 1] + dp[i - coins[j - 1]][j]
                } else {
                    dp[i][j] = dp[i][j - 1]
                }
            }
        }
        
        return dp[a - 1][c - 1]
    }
}