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?微信公眾號：山青詠芝（shanqingyongzhi）
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?原文地址：https://www.cnblogs.com/strengthen/p/10397495.html?
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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:

All the given strings' lengths will not exceed 10.

The length of the given list will be in the range of [2, 50].

---

給定字符串列表，你需要從它們中找出最長的特殊序列。最長特殊序列定義如下：該序列為某字符串獨有的最長子序列（即不能是其他字符串的子序列）。

子序列可以通過刪去字符串中的某些字符實現，但不能改變剩余字符的相對順序?？招蛄袨樗凶址淖有蛄?#xff0c;任何字符串為其自身的子序列。

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

示例：

輸入: "aba", "cdc", "eae" 輸出: 3?

提示：

所有給定的字符串長度不會超過 10 。

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函數可以檢索數組中的值 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 }

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轉載于:https://www.cnblogs.com/strengthen/p/10397495.html

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