[Swift]LeetCode522. 最長特殊序列 II | Longest Uncommon Subsequence II

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Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be a list of strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1.

Example 1:

Input: "aba", "cdc", "eae"
Output: 3 

Note:

1. All the given strings' lengths will not exceed 10.
2. The length of the given list will be in the range of [2, 50].

---

給定字符串列表，你須要從它們中找出最長的特殊序列。最長特殊序列定義以下：該序列爲某字符串獨有的最長子序列（即不能是其餘字符串的子序列）。

子序列能夠經過刪去字符串中的某些字符實現，但不能改變剩餘字符的相對順序。空序列爲全部字符串的子序列，任何字符串爲其自身的子序列。

輸入將是一個字符串列表，輸出是最長特殊序列的長度。若是最長特殊序列不存在，返回 -1 。 

示例：

輸入: "aba", "cdc", "eae"
輸出: 3 

提示：

1. 全部給定的字符串長度不會超過 10 。
2. 給定字符串列表的長度將在 [2, 50 ] 之間。

---

Runtime: 16 ms

Memory Usage: 19.8 MB

 1 class Solution {
 2     func findLUSlength(_ strs: [String]) -> Int {
 3         var strs = strs
 4         var n:Int = strs.count
 5         var s:Set<String> = Set<String>()
 6         strs.sort(by:{
 7             if $0.count == $1.count
 8             {
 9                 return $0 > $1
10             }
11              return $0.count > $1.count
12         })
13         for i in 0..<n
14         {
15             if i == n - 1 || strs[i] != strs[i + 1]
16             {
17                 var found:Bool = true
18                 for a in s
19                 {
20                     var j:Int = 0
21                     for c in a.characters
22                     {
23                         if c == strs[i][j] {j += 1}
24                         if j == strs[i].count {break}
25                     }
26                     if j == strs[i].count
27                     {
28                         found = false
29                         break
30                     }
31                 }
32                 if found {return strs[i].count}                
33             }
34             s.insert(strs[i])
35         }
36         return -1
37     }
38 }
39 
40 extension String {        
41     //subscript函數能夠檢索數組中的值
42     //直接按照索引方式截取指定索引的字符
43     subscript (_ i: Int) -> Character {
44         //讀取字符
45         get {return self[index(startIndex, offsetBy: i)]}
46     }
47 }

---

16ms

 1 class Solution {
 2     func findLUSlength(_ strs: [String]) -> Int {
 3         var sortStrs = strs.sorted { 
 4             return $1.count < $0.count || ($1.count == $0.count && $1 < $0 )
 5         }
 6         let duplicates = getDuplicates(sortStrs)
 7         for i in sortStrs.indices {
 8             if !duplicates.contains(sortStrs[i]) {
 9                 if i == 0 { return sortStrs[i].count }
10                 for j in 0..<i {
11                     if isSubsequence(Array(sortStrs[j]), Array(sortStrs[i])) {
12                         break
13                     }
14                     if j == i-1 {
15                         return sortStrs[i].count
16                     }
17                 }
18             }
19         }
20         return -1
21     }
22 
23     func isSubsequence(_ chars1: [Character], _ chars2: [Character]) -> Bool {
24         var i = 0, j = 0
25         while i<chars1.count && j<chars2.count {
26             if chars1[i] == chars2[j] {
27                 j += 1
28             }
29             i += 1
30         }
31         return j == chars2.count
32     }
33 
34     func getDuplicates(_ strs: [String]) -> Set<String> {
35         var set = Set<String>()
36         var duplicates = Set<String>()
37         for s in strs {
38             if set.contains(s) {
39                 duplicates.insert(s)
40             }else {
41                 set.insert(s)
42             }
43         }
44         return duplicates
45     }
46 }