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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