[Swift]LeetCode214. 最短回文串 | Shortest Palindrome
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10197925.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.
Example 1:
Input:"aacecaaa"Output:"aaacecaaa"
Example 2:
Input:"abcd"Output:"dcbabcd"
给定一个字符串 s，你可以通过在字符串前面添加字符将其转换为回文串。找到并返回可以用这种方式转换的最短回文串。
示例 1:
输入:"aacecaaa"输出:"aaacecaaa"
示例 2:
输入:"abcd"输出:"dcbabcd"
52ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 let characters = Array(s) 4 var end = s.count 5 while true { 6 var j = 0 7 var i = end-1 8 while i >= 0 { 9 if characters[i] == characters[j] { 10 j += 1 11 } 12 i -= 1 13 } 14 if j == end { 15 break 16 } 17 end = j 18 } 19 return (s.suffix(s.count-end).reversed() + s) 20 } 21 }
60ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 let sArray = Array(s) 4 var j = 0 5 for i in (0..<s.count).reversed() { 6 if sArray[i] == sArray[j] { 7 j += 1 8 } 9 } 10 if j == s.count { return s } 11 let suffix = String(s[s.index(s.startIndex, offsetBy: j)...]) 12 let prefix = suffix.reversed() 13 let mid = shortestPalindrome(String(s[s.index(s.startIndex, offsetBy:0)..<s.index(s.startIndex, offsetBy:j)])) 14 15 return prefix + mid + suffix 16 } 17 }
68ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 return String(checkShortest(Array(s))) 4 } 5 6 func checkShortest(_ s: [Character]) -> [Character] { 7 if s.count <= 1 { 8 return s 9 } 10 11 var j = 0 12 for i in stride(from: s.count - 1, through: 0, by: -1) { 13 if s[s.index(s.startIndex, offsetBy: i)] == s[s.index(s.startIndex, offsetBy: j)] { 14 j += 1 15 } 16 } 17 18 if j == s.count { 19 return s 20 } 21 22 let suffix = s[j..<s.count] 23 let middle = checkShortest(Array(s[0..<j])) 24 let prefix = suffix.reversed() 25 return prefix + middle + suffix 26 } 27 }
80ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 let sArray = Array(s) 4 var j = 0 5 for i in (0..<sArray.count).reversed() { 6 if sArray[i] == sArray[j] { 7 j += 1 8 } 9 } 10 if j == sArray.count { return s } 11 let suffix = String(sArray[j...]) 12 let prefix = suffix.reversed() 13 let mid = shortestPalindrome(String(sArray[0..<j])) 14 return prefix + mid + suffix 15 } 16 }
84ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 if s.isEmpty { 4 return s 5 } 6 7 var sArr = Array(s) 8 9 if sArr == sArr.reversed() { 10 return s 11 } 12 13 var mArr = [Character]() 14 for i in 0..<sArr.count { 15 mArr.append("#") 16 mArr.append(sArr[i]) 17 } 18 mArr.append("#") 19 var counts = Array(repeating: 0, count: mArr.count) 20 var right = 1 21 var center = 1 22 var left = sArr.count - 1 23 for i in 1..<mArr.count / 2 { 24 let minx = right <= i ? 1 : min(right - i, counts[center * 2 - i] + 1) 25 for r in minx...i { 26 if mArr[i-r] != mArr[i+r] { 27 break 28 } 29 counts[i] = r 30 if i==r { 31 left = sArr.count - i 32 } 33 } 34 35 if right < i + counts[i] { 36 right = i + counts[i] 37 center = i 38 } 39 } 40 var res = [Character]() 41 var k = sArr.count - 1 42 for _ in 0..<left { 43 res.append(sArr[k]) 44 k -= 1 45 } 46 res.append(contentsOf: sArr) 47 48 return String(res) 49 } 50 }
128ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 let characters = Array(s) 4 var end = s.count 5 while true { 6 var j = 0 7 var i = end-1 8 while i >= 0 { 9 if characters[i] == characters[j] { 10 j += 1 11 } 12 i -= 1 13 } 14 if j == end { 15 break 16 } 17 end = j 18 } 19 let prefix = String(s.suffix(s.count-end).reversed()) 20 return (prefix + s) 21 } 22 }
3008ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 var i:Int = 0 4 var end:Int = s.count - 1 5 var j:Int = end 6 var arr:[Character] = [Character]() 7 for char in s.characters 8 { 9 arr.append(char) 10 } 11 while (i < j) 12 { 13 if arr[i] == arr[j] 14 { 15 i += 1 16 j -= 1 17 } 18 else 19 { 20 i = 0 21 end -= 1 22 j = end 23 } 24 } 25 let res:String = s.subString(end + 1).reversed() + s 26 return res 27 } 28 } 29 30 extension String 31 { 32 // 截取字符串：从index到结束处 33 // - Parameter index: 开始索引 34 // - Returns: 子字符串 35 func subString(_ index: Int) -> String { 36 let theIndex = self.index(self.endIndex, offsetBy: index - self.count) 37 return String(self[theIndex..<endIndex]) 38 } 39 }
4052ms
1 class Solution { 2 func shortestPalindrome(_ s: String) -> String { 3 let characters = Array(s) 4 let count = s.count 5 var end = count-1 6 while end >= 1 { 7 if !isPalindrome(arr: characters, end: end) { 8 end -= 1 9 } else { 10 break 11 } 12 } 13 // let prefix = String(s.suffix(count-1-end).reversed()) 14 return (s.suffix(count-1-end).reversed() + s) 15 } 16 17 func isPalindrome(arr: [Character], end: Int) -> Bool { 18 for i in 0 ... end/2 { 19 let endIndex = end-i 20 if arr[i] != arr[endIndex] { 21 return false 22 } 23 } 24 return true 25 } 26 }
