[Swift]LeetCode93. 復原IP地址 | Restore IP Addresses

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➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-ufrnjkux-me.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
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Given a string containing only digits, restore it by returning all possible valid IP address combinations.

Example:

Input: "25525511135"
Output: ["255.255.11.135", "255.255.111.35"]

---

給定一個只包含數字的字符串，復原它並返回全部可能的 IP 地址格式。

示例:

輸入: "25525511135"
輸出: ["255.255.11.135", "255.255.111.35"]

---

16ms

 1 class Solution {
 2     func restoreIpAddresses(_ s: String) -> [String] {
 3         if s.characters.count < 4 || s.characters.count > 12 {
 4             return []
 5         }
 6         
 7         let characters = Array(s.characters)
 8         var result = [String]()
 9         var candidate = [String]()
10         
11         backtracking(characters, 0, &candidate, &result)
12         
13         return result
14     }
15     private func backtracking(_ characters: [Character], _ pos: Int, _ candidate: inout [String], _ result: inout [String]) {
16         if candidate.count == 4 {
17             result.append(candidate.joined(separator: "."))
18             return
19         }
20         
21         let charsLeft = characters.count - pos
22         let groupsLeft = 4 - candidate.count
23         let minLen = groupsLeft == 1 ? charsLeft - groupsLeft + 1 : 1
24         let maxLen = characters[pos] == "0" ? 1 : min(3, charsLeft - groupsLeft + 1)
25         
26         if minLen > maxLen {
27             return
28         }
29 
30         for len in minLen...maxLen {
31             let num = String(characters[pos..<(pos + len)])
32             if Int(num)! > 255 {
33                 continue
34             }
35             candidate.append(num)
36             backtracking(characters, pos + len, &candidate, &result)
37             candidate.removeLast()
38         }
39     }
40 }