[Swift]LeetCode49. 字母異位詞分組 | Group Anagrams

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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-bvnfejbb-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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array of strings, group anagrams together.

Example:

Input: ,
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]["eat", "tea", "tan", "ate", "nat", "bat"]

Note:

• All inputs will be in lowercase.
• The order of your output does not matter.

---

給定一個字符串數組，將字母異位詞組合在一塊兒。字母異位詞指字母相同，但排列不一樣的字符串。

示例:

輸入: ,
輸出:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]["eat", "tea", "tan", "ate", "nat", "bat"]

說明：

• 全部輸入均爲小寫字母。
• 不考慮答案輸出的順序。

---

68ms

 1 class Solution {
 2     func hashValue(_ str: String) -> UInt64 {
 3         var hash: UInt64 = 0
 4         for code in str.unicodeScalars {
 5             let value = code.value - 97
 6             hash += UInt64(pow(5.0, Double(value)))
 7         }
 8 
 9         return hash
10     }
11 
12     func groupAnagrams(_ strs: [String]) -> [[String]] {
13         var dic = [UInt64 : [String]]()
14         for str in strs {
15             let hash = hashValue(str)
16             if let strings = dic[hash] {
17                 dic[hash] = strings + [str]
18             } else {
19                 dic[hash] = [str]
20             }
21         }
22         return Array(dic.values)
23     }
24 }

---

124ms

 1 class Solution {
 2     func groupAnagrams(_ strs: [String]) -> [[String]] {
 3         var result = [String : [String]]()
 4         for item in strs {
 5             let newStr = reformateString(item)
 6             if let tempArray = result[newStr] {
 7                 var newArray = tempArray
 8                 newArray.append(item)
 9                 result[newStr] = newArray
10             } else {
11                 result[newStr] = [item]
12             }
13         }
14         return [[String]](result.values)
15     }
16     
17     func reformateString(_ str: String) -> String {
18        return String([Character](str).sorted(by: <))
19     }
20 }

---

132ms

 1 class Solution {
 2     func groupAnagrams(_ strs: [String]) -> [[String]] {
 3         var res = [[String]]()
 4         var dict = [String: [String]]()
 5         strs.forEach{(string) in
 6             let charArray = Array(string).sorted(){$0 < $1}
 7             let key = String(charArray)
 8             if dict[key] != nil {
 9                 dict[key]?.append(string)
10             }else {
11                 dict[key] = [string]
12             }
13         }
14         for key in dict.keys {
15             res.append(dict[key]!)
16         }
17         return res
18     }
19 }

---

140ms

 1 class Solution {
 2     func groupAnagrams(_ strs: [String]) -> [[String]] {
 3         var map = [String: [String]]()
 4         
 5         for str in strs {
 6             let key = String(str.sorted())
 7             map[key, default: []].append(str)
 8         }
 9         
10         return Array(map.values)
11     }
12 }

---

148ms

 1 class Solution {
 2     func groupAnagrams(_ strs: [String]) -> [[String]] {
 3         var map: [String: [String]] = [:]
 4         for str in strs {
 5             let chars = str.sorted()
 6             let key = String(chars)
 7             if var list = map[key] {
 8                 list.append(str)
 9                 map[key] = list
10             } else {
11                 map[key] = [str]
12             }
13         }
14         return Array(map.values)
15     }
16 }

---

316ms

 1 class Solution {
 2     func groupAnagrams(_ strs: [String]) -> [[String]] {
 3         var groups = [String: [String]]()
 4         for str in strs {
 5             let strSorted = String(str.sorted())
 6             if groups[strSorted] == nil {
 7                 groups[strSorted] = [str]
 8             } else {
 9                 groups[strSorted]?.append(str)
10             }
11         }
12         return groups.map { $1 }
13     }
14 }