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：
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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?
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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：
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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)
}
}
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之后就可以利用上面的算法去完成前面说的两次翻转：
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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)
}
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完整算法代码：
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//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"
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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，谢谢大家对服务器之家的支持。
原文链接：http://www.imlifengfeng.com/blog/?p=643