小编给大家分享一下编程语言中无重复字符最长子串的示例分析，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！
给定一个字符串，找出不含有重复字符的 最长子串 的长度。
示例：
给定 `"abcabcbb"` ，没有重复字符的最长子串是 `"abc"` ，那么长度就是3。
给定 `"bbbbb"` ，最长的子串就是 `"b"` ，长度是1。
给定 `"pwwkew"` ，最长子串是 `"wke"` ，长度是3。请注意答案必须是一个 子串 ， `"pwke"` 是 _子序列_ 而不是子串。
1. Swift 语言
class LongestSubstringWithoutRepeatingCharacters {
func lengthOfLongestSubstring(_ s: String) -> Int {
var longest = 0, left = 0, set = Set<Character>()
let sChars = Array(s)
for (i, char) in sChars.enumerated() {
if set.contains(char) {
longest = max(longest, i - left)
while sChars[left] != char {
set.remove(sChars[left])
left += 1
}
left += 1
} else {
set.insert(char)
}
}
return max(longest, sChars.count - left)
}
}2. JavaScript 语言
/**
* @param {string} s
* @return {number}
*/
var lengthOfLongestSubstring = function(s) {
var hash = {};
var start = 0;
var ans = 0;
for (var i = 0, len = s.length; i < len; i++) {
var item = s[i];
if (!hash[item])
hash[item] = true;
else {
// item 已经在 substring 中存在了
for (; ;) {
if (s[start] === item) {
start++;
break;
}
hash[s[start]] = false;
start++;
}
}
ans = Math.max(ans, i - start + 1);
}
return ans;
};3. Python 语言
class Solution(object):
def _lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
d = collections.defaultdict(int)
l = ans = 0
for i, c in enumerate(s):
while l > 0 and d[c] > 0:
d[s[i-l]] -= 1
l -= 1
d[c] += 1
l += 1
ans = max(ans, l)
return ans
def lengthOfLongestSubstring(self, s):
d = {}
start = 0
ans = 0
for i,c in enumerate(s):
if c in d:
start = max(start, d[c] + 1)
d[c] = i
ans = max(ans, i - start + 1)
return ans4. Java 语言
// 方式一
public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length();
int ans = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j <= n; j++)
if (allUnique(s, i, j)) ans = Math.max(ans, j - i);
return ans;
}
public boolean allUnique(String s, int start, int end) {
Set<Character> set = new HashSet<>();
for (int i = start; i < end; i++) {
Character ch = s.charAt(i);
if (set.contains(ch)) return false;
set.add(ch);
}
return true;
}
}
// 方式二
public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length();
Set<Character> set = new HashSet<>();
int ans = 0, i = 0, j = 0;
while (i < n && j < n) {
// try to extend the range [i, j]
if (!set.contains(s.charAt(j))){
set.add(s.charAt(j++));
ans = Math.max(ans, j - i);
}
else {
set.remove(s.charAt(i++));
}
}
return ans;
}
}
// 方式三
public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length(), ans = 0;
Map<Character, Integer> map = new HashMap<>(); // current index of character
// try to extend the range [i, j]
for (int j = 0, i = 0; j < n; j++) {
if (map.containsKey(s.charAt(j))) {
i = Math.max(map.get(s.charAt(j)), i);
}
ans = Math.max(ans, j - i + 1);
map.put(s.charAt(j), j + 1);
}
return ans;
}
}5. C++ 语言
#include <string.h>
#include <iostream>
#include <string>
#include <map>
using namespace std;
int lengthOfLongestSubstring1(string s) {
map<char, int> m;
int maxLen = 0;
int lastRepeatPos = -1;
for(int i=0; i<s.size(); i++){
if (m.find(s[i])!=m.end() && lastRepeatPos < m[s[i]]) {
lastRepeatPos = m[s[i]];
}
if ( i - lastRepeatPos > maxLen ){
maxLen = i - lastRepeatPos;
}
m[s[i]] = i;
}
return maxLen;
}
//don't use <map>
int lengthOfLongestSubstring(string s) {
const int MAX_CHARS = 256;
int m[MAX_CHARS];
memset(m, -1, sizeof(m));
int maxLen = 0;
int lastRepeatPos = -1;
for(int i=0; i<s.size(); i++){
if (m[s[i]]!=-1 && lastRepeatPos < m[s[i]]) {
lastRepeatPos = m[s[i]];
}
if ( i - lastRepeatPos > maxLen ){
maxLen = i - lastRepeatPos;
}
m[s[i]] = i;
}
return maxLen;
}
int main(int argc, char** argv)
{
string s = "abcabcbb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
s = "bbbbb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
s = "bbabcdb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
if (argc>1){
s = argv[1];
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
}
return 0;
}6. C 语言
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static int lengthOfLongestSubstring(char *s)
{
int offset[128];
int max_len = 0;
int len = 0;
int index = 0;
memset(offset, 0xff, sizeof(offset));
while (*s != '\0') {
if (offset[*s] == -1) {
len++;
} else {
if (index - offset[*s] > len) {
len++;
} else {
len = index - offset[*s];
}
}
if (len > max_len) {
max_len = len;
}
offset[*s++] = index++;
}
return max_len;
}
int main(int argc, char **argv)
{
if (argc != 2) {
fprintf(stderr, "Usage: ./test string
");
exit(-1);
}
printf("%d
", lengthOfLongestSubstring(argv[1]));
return 0;
}以上是“编程语言中无重复字符最长子串的示例分析”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注亿速云行业资讯频道！
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