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Swift算法实现逐字翻转字符串的方法示例

作者:李峰峰博客

大家都知道翻转字符串在字符串算法中算是比较常见的,下面这篇文章主要介绍了Swift算法实现逐字翻转字符串的方法,文中给出了详细的示例代码,需要的朋友可以参考借鉴,下面来一起看看吧。

前言

翻转字符串在字符串算法中算是比较常见的,而且被很多公司用作笔试题。”逐字翻转字符串”是翻转字符串的翻版,也是之前Google的面试题,原题是这样的:

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?

简而言之就是:”the sky is blue”—>”blue is sky the”

所以,对于本文,要解决的算法是:

逐字翻转字符串,例如:"the sky is blue"—>"blue is sky the"

接下来看下实现思路和代码。

实现思路及代码

既然是字符串翻转的翻版,我们就可以利用之前翻版字符串的思路去解决就可以了,不过这道题要有两次翻转:

第一次翻转,整体翻转:”the sky is blue” -> “eulb si yks eht”

第二次翻转,单词翻转:”eulb si yks eht” -> “blue is sky the”

所以,首先可以实现一个可以翻转局部和全部字符串的算法,传入字符数组、startIndex 和 endIndex ,其中 startIndex 和 endIndex 分别为要翻转的字符串的起始下标和结束下标,也就是要翻转 startIndex 和 endIndex 之间(包含)的字符,代码如下:

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说的两次翻转:

func reverseWords(_ str:String) -> String{
 
 var chars = [Character](str.characters)
 
 //首先翻转整个字符串所有字符,"the sky is blue" -> "eulb si yks eht"
 _reverseStr(&chars,0,chars.count-1)
 
 //然后翻转每个单词中的字符,"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)
}

完整算法代码:

//翻转指定范围的字符
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)
  
 }
 
}
 
//逐字翻转字符串
func reverseWords(_ str:String) -> String{
 
 var chars = [Character](str.characters)
 
 //首先翻转整个字符串所有字符,"the sky is blue" -> "eulb si yks eht"
 _reverseStr(&chars,0,chars.count-1)
 
 //然后翻转每个单词中的字符,"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"

总结

以上就是关于Swift算法实现逐字翻转字符串的方法,希望本文的内容对大家的学习或者工作能带来一定的帮助,如果有疑问大家可以留言交流,谢谢大家对脚本之家的支持。

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