[Swift]LeetCode518.零钱兑换II|CoinChange2

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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

 1 class Solution {
 2     func change(_ amount: Int, _ coins: [Int]) -> Int {
 3         var memo = Array(repeating: 0, count: amount + 1)
 4         memo[0] = 1
 5         
 6         var coins = coins.sorted()
 7         
 8         for coin in coins where coin <= amount {
 9             for higherAmount in coin...amount {
10                 memo[higherAmount] += memo[higherAmount - coin]
11             }
12         }
13         
14         return memo[amount]
15     }
16 }

184ms

 1 class Solution {
 2     var cache: [Int: Int] = [:]
 3     func change(_ amount: Int, _ coins: [Int]) -> Int {
 4         return helper(amount, 0, coins.sorted(by: >))
 5     }
 6     
 7     private func helper(_ amount: Int, _ index: Int, _ coins: [Int]) -> Int {
 8         if amount == 0 { 
 9             return 1 
10         }
11         if index >= coins.count { return 0 }
12         if let some = cache[5000 * index + amount] {
13             return some
14         }
15         
16         var res = 0
17         for i in index..<coins.count {
18             let coin = coins[i]
19             if coin <= amount {
20                 res += helper(amount - coin, i, coins)
21             }
22         }
23         cache[5000 * index + amount] = res
24         return res
25     }
26 }

316ms

 1 class Solution {
 2     func change(_ amount: Int, _ coins: [Int]) -> Int {
 3         var row = [1]
 4         row.append(contentsOf: [Int](repeating: 0, count: amount))
 5         var dp = [[Int]](repeating: row, count: coins.count + 1)
 6         // i: number of coins, the last index in this range is i - 1
 7         guard 1 <= coins.count else {
 8             return dp[coins.count][amount]
 9         }
10         for i in 1...coins.count {
11             guard 1 <= amount else {
12                 continue
13             }
14             for j in 1...amount {
15                 dp[i][j] = dp[i - 1][j] + (j < coins[i - 1] ? 0 : dp[i][j - coins[i - 1]])
16             }
17         }
18         return dp[coins.count][amount]
19     }
20 }

332ms

 1 class Solution {
 2     func change(_ amount: Int, _ coins: [Int]) -> Int {
 3         let a = amount + 1
 4         let c = coins.count + 1
 5         
 6         var dp = [[Int]](repeating: [Int](repeating: 0, count: c), count: a)
 7         
 8         for j in 0..<c {
 9             dp[0][j] = 1
10         }
11         
12         for i in 1..<a {
13             for j in 1..<c {
14                 if coins[j - 1] <= i {
15                     dp[i][j] = dp[i][j - 1] + dp[i - coins[j - 1]][j]
16                 } else {
17                     dp[i][j] = dp[i][j - 1]
18                 }
19             }
20         }
21         
22         return dp[a - 1][c - 1]
23     }
24 }