前言
翻轉字符串在字符串算法中算是比較常見的，而且被很多公司用作筆試題。”逐字翻轉字符串”是翻轉字符串的翻版，也是之前Google的面試題，原題是這樣的：
Given an input string, reverse the string word by word.A word is defined as a sequence of non-space characters.The input string does not contain leading or trailing spaces and the words are always separated by a single space.For example,Given s = "the sky is blue",return "blue is sky the".Could you do it in-place without allocating extra space?
簡而言之就是：”the sky is blue”—>”blue is sky the”
所以，對于本文，要解決的算法是：
逐字翻轉字符串，例如："the sky is blue"—>"blue is sky the"
接下來看下實現思路和代碼。
實現思路及代碼
既然是字符串翻轉的翻版，我們就可以利用之前翻版字符串的思路去解決就可以了，不過這道題要有兩次翻轉：
第一次翻轉，整體翻轉：”the sky is blue” -> “eulb si yks eht”
第二次翻轉，單詞翻轉：”eulb si yks eht” -> “blue is sky the”
所以，首先可以實現一個可以翻轉局部和全部字符串的算法，傳入字符數組、startIndex 和 endIndex ，其中 startIndex 和 endIndex 分別為要翻轉的字符串的起始下標和結束下標，也就是要翻轉 startIndex 和 endIndex 之間（包含）的字符，代碼如下：
func _reverseStr( _ chars:inout [Character], _ startIndex:Int, _ endIndex:Int){ var startIndex = startIndex var endIndex = endIndex if startIndex <= endIndex { let tempChar = chars[endIndex] chars[endIndex] = chars[startIndex] chars[startIndex] = tempChar startIndex += 1 endIndex -= 1 _reverseStr(&chars,startIndex,endIndex) } }之后就可以利用上面的算法去完成前面說的兩次翻轉：
func reverseWords(_ str:String) -> String{ var chars = [Character](str.characters) //首先翻轉整個字符串所有字符,"the sky is blue" -> "eulb si yks eht" _reverseStr(&chars,0,chars.count-1) //然后翻轉每個單詞中的字符，"eulb si yks eht" -> "blue is sky the" var startIndex = 0 for endIndex in 0 ..< chars.count { if endIndex == chars.count - 1 || chars[endIndex + 1] == " " { _reverseStr(&chars, startIndex, endIndex) startIndex = endIndex + 2 } } return String(chars)}完整算法代碼：
//翻轉指定范圍的字符func _reverseStr( _ chars:inout [Character], _ startIndex:Int, _ endIndex:Int){ var startIndex = startIndex var endIndex = endIndex if startIndex <= endIndex { let tempChar = chars[endIndex] chars[endIndex] = chars[startIndex] chars[startIndex] = tempChar startIndex += 1 endIndex -= 1 _reverseStr(&chars,startIndex,endIndex) } } //逐字翻轉字符串func reverseWords(_ str:String) -> String{ var chars = [Character](str.characters) //首先翻轉整個字符串所有字符,"the sky is blue" -> "eulb si yks eht" _reverseStr(&chars,0,chars.count-1) //然后翻轉每個單詞中的字符，"eulb si yks eht" -> "blue is sky the" var startIndex = 0 for endIndex in 0 ..< chars.count { if endIndex == chars.count - 1 || chars[endIndex + 1] == " " { _reverseStr(&chars, startIndex, endIndex) startIndex = endIndex + 2 } } return String(chars)} reverseWords("the sky is blue") //return "blue is sky the"總結
以上就是關于Swift算法實現逐字翻轉字符串的方法，希望本文的內容對大家的學習或者工作能帶來一定的幫助，如果有疑問大家可以留言交流，謝謝大家對VEVB武林網的支持。
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