[LeetCode3] Longest Substring Without Repeating Characters

最长无重复子串算法

最新推荐文章于 2025-11-24 01:23:29 发布

原创最新推荐文章于 2025-11-24 01:23:29 发布 · 869 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #leetcode #hashmap

Leetcode 同时被 2 个专栏收录

287 篇文章

订阅专栏

LeetCode Questions

178 篇文章

订阅专栏

本文介绍了一种寻找字符串中最长无重复字符子串的方法，并提供了C++和Java两种实现方式。通过从左到右扫描字符串，遇到重复字符时更新搜索起点，最终找到最长无重复字符子串的长度。

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

从左往右扫描，当遇到重复字母时，以上一个重复字母的index +1，作为新的搜索起始位置。比如

直到扫描到最后一个字母。
C++实现

<span style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;color:#333333;">int lengthOfLongestSubstring(string s) {
        int maxL=0;
        int len =0;
        int counts[26];
        memset(counts, -1, 26*sizeof(int));
    
        for(int i=0; i<s.size(); i++,len++){
            if(counts[s[i]-'a']>=0){
            maxL = max(maxL, len);
            len = 0;
            i = counts[s[i]-'a']+1;
            memset(counts,-1,26*sizeof(int));
        }
        counts[s[i]-'a'] = i;
    }
    return max(len, maxL);
    }</span>

Java实现

<span style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;color:#333333;">public int lengthOfLongestSubstring(String s) {
        if(s.length() == 0) return 0;
		int start = 0;
		int end = 0;
		int maxLen = 0;
		HashMap<Character, Integer> charMap = new HashMap<Character,Integer>();
		while(end<s.length()){
			if(!charMap.containsKey(s.charAt(end))){
				charMap.put(s.charAt(end), end);
				int len = end-start+1;
				maxLen = Math.max(len, maxLen);
			}else{
				int val = charMap.get(s.charAt(end));
				for(int i=start;i<val+1;i++)
					charMap.remove(s.charAt(i));
				start = val+1;
				charMap.put(s.charAt(end), end);
			}
			end++;
		}
		return maxLen;
    }</span>