Swift算法實現(xiàn)逐字翻轉(zhuǎn)字符串的方法示例
前言
翻轉(zhuǎn)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(zhuǎn)字符串”是翻轉(zhuǎn)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(zhuǎn)字符串,例如:"the sky is blue"—>"blue is sky the"
接下來看下實現(xiàn)思路和代碼。
實現(xiàn)思路及代碼
既然是字符串翻轉(zhuǎn)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(zhuǎn):
第一次翻轉(zhuǎn),整體翻轉(zhuǎn):”the sky is blue” -> “eulb si yks eht”
第二次翻轉(zhuǎn),單詞翻轉(zhuǎn):”eulb si yks eht” -> “blue is sky the”
所以,首先可以實現(xiàn)一個可以翻轉(zhuǎn)局部和全部字符串的算法,傳入字符數(shù)組、startIndex 和 endIndex ,其中 startIndex 和 endIndex 分別為要翻轉(zhuǎn)的字符串的起始下標和結(jié)束下標,也就是要翻轉(zhuǎn) 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(zhuǎn):
func reverseWords(_ str:String) -> String{
var chars = [Character](str.characters)
//首先翻轉(zhuǎn)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(zhuǎn)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(zhuǎn)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(zhuǎn)字符串
func reverseWords(_ str:String) -> String{
var chars = [Character](str.characters)
//首先翻轉(zhuǎn)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(zhuǎn)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(jié)
以上就是關于Swift算法實現(xiàn)逐字翻轉(zhuǎn)字符串的方法,希望本文的內(nèi)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,謝謝大家對腳本之家的支持。
相關文章
Swift中的條件切換語句switch...case學習教程
這篇文章主要介紹了Swift中的條件切換語句switch...case學習教程,Swift中的switch...case支持的數(shù)據(jù)類型很多,非常之強大,需要的朋友可以參考下2016-04-04

