查找不重复字符的最长子字符串（编程面试中常见题-用8种编程语言来回答）
给定一个字符串str，找到不重复字符的最长子字符串。
比如我们有 “ABDEFGABEF”, 最长的字符串是 “BDEFGA” 和 “DEFGAB”, 长度为6.
再如 “BBBB” 最长字符串是 “B”, 长度为1.
再比如 “neatcoding” 最长字符串是“neatcodi”, 长度为8.
所需的时间复杂度为O（n），其中n是字符串的长度。
我们将逐个字符地遍历该字符串
检查此字符是否在当前子字符串中，如果是，则将当前子字符串保存到子字符串集合中，使用此字符作为起始值创建一个新的子字符串；
如果否，则将此字符添加到当前子字符串中；
至此，循环结束，检查当前子字符串是否为空，如果是，则不执行任何操作
如果否，请将其添加到子字符串集合
我们设置一个对象来存储最长的子字符串，
现在，让我们遍历子字符串集合，找到最长的一个
检查当前子字符串是否更长，如果是，则替换最长的字符串
如果没有，什么也不做
最后，我们有最长的子字符串。
让我们编码。
InJavascript:
returnLongestNonRepeatSubString(s){
if(!s){
return "";
}
let subCollection = [];
let currentSub = "";
for(let i = 0; i < s.length; i ++) {
let c = s[i];
if(currentSub.indexOf(c) === -1) {
currentSub += c;
}
else {
subCollection.push(currentSub);
currentSub = "" + c;
}
}
if(currentSub.length > 0){
subCollection.push(currentSub);
}
let longest = "";
for(let i = 0 ; i < subCollection.length ; i ++) {
let sub = subCollection[i];
if(sub.length > longest.length) {
longest = sub;
}
}
return longest;
}
InC#:
string returnLongestNonRepeatSubString(string s)
{
if (string.IsNullOrEmpty(s))
{
return "";
}
List<string> subCollection = new List<string>();
string currentSub = "";
for (int i = 0; i < s.Length; i++)
{
char c = s[i];
if (currentSub.IndexOf(c) == -1)
{
currentSub += c;
}
else
{
subCollection.Add(currentSub);
currentSub = "" + c;
}
}
if (currentSub.Length > 0)
{
subCollection.Add(currentSub);
}
string longest = "";
for (int i = 0; i < subCollection.Count; i++)
{
string sub = subCollection[i];
if (sub.Length > longest.Length)
{
longest = sub;
}
}
return longest;
}
InJava:
String returnLongestNonRepeatSubString(String s)
{
if (s == null || s.length() == 0)
{
return "";
}
List<String> subCollection = new ArrayList<String>();
String currentSub = "";
for (int i = 0; i < s.length(); i++)
{
char c = s.charAt(i);
if (currentSub.indexOf(c) == -1)
{
currentSub += c;
}
else
{
subCollection.add(currentSub);
currentSub = "" + c;
}
}
if (currentSub.length() > 0)
{
subCollection.add(currentSub);
}
String longest = "";
for (int i = 0; i < subCollection.size(); i++)
{
String sub = subCollection.get(i);
if (sub.length() > longest.length())
{
longest = sub;
}
}
return longest;
}
InKotlin:
fun returnLongestNonRepeatSubString(s: String?): String {
if (s == null || s.length == 0) {
return ""
}
val subCollection = ArrayList<String>()
var currentSub = ""
for (i in 0 until s.length) {
val c = s[i]
if (currentSub.indexOf(c) == -1) {
currentSub += c
} else {
subCollection.add(currentSub)
currentSub = "" + c
}
}
if (currentSub.length > 0) {
subCollection.add(currentSub)
}
var longest = ""
for (i in subCollection.indices) {
val sub = subCollection[i]
if (sub.length > longest.length) {
longest = sub
}
}
return longest
}
InGolang:
func returnLongestNonRepeatSubString(s string) string {
if len(s) == 0 {
return ""
}
subCollection := make([]string, 1)
currentSub := ""
for i := 0; i < len(s); i++ {
var c = s[i]
if !strings.Contains(currentSub, string(c)) {
currentSub += string(c)
} else {
subCollection = append(subCollection, currentSub)
currentSub = string(c)
}
}
if len(currentSub) > 0 {
subCollection = append(subCollection, currentSub)
}
var longest = ""
for _, sub := range subCollection {
if len(sub) > len(longest) {
longest = sub
}
}
return longest
}
In c++:
string
returnLongestNonRepeatSubString (const string & s)
{
if (s.empty ())
{
return "";
}
std::vector <string> subCollection;
string currentSub;
for (int i = 0; i < s.length (); i++)
{
char c = s[i];
if (currentSub.find (c) == -1)
{
currentSub += c;
}
else
{
subCollection.push_back (currentSub);
currentSub = string (1, c);
}
}
if(!currentSub.empty())
{
subCollection.push_back (currentSub);
}
string longest = "";
for (int i = 0; i < subCollection.size(); i++)
{
string sub = subCollection.at(i);
if (sub.length() > longest.length())
{
longest = sub;
}
}
return longest;
}
InPython:
def returnLongestNonRepeatSubString(s):
if not s:
return ""
subCollection = []
currentSub = ""
for c in s:
if (currentSub.find(c) == -1):
currentSub += c
else:
subCollection.append(currentSub)
currentSub = c
if currentSub:
subCollection.append(currentSub)
longest = ""
for sub in subCollection:
if (len(sub) > len(longest)):
longest = sub
return longest
InSwift:
func returnLongestNonRepeatSubString (s: String) -> String {
if (s == nil || s.isEmpty) {
return ""
}
var subCollection = [String]()
var currentSub = ""
for c in s {
let i = currentSub.firstIndex(of: c)
if(i == nil) {
currentSub = currentSub + String(c)
}
else {
subCollection.append(currentSub);
currentSub = String(c)
}
}
var longest = ""
for sub in subCollection {
if sub.length > longest.length {
longest = sub
}
}
return longest
}
