LeetCode 3. Longest Substring Without Repeating Characters

最新推荐文章于 2021-12-03 15:58:56 发布

原创最新推荐文章于 2021-12-03 15:58:56 发布 · 153 阅读

0 ·

CC 4.0 BY-SA版权

Algorithm 同时被 2 个专栏收录

15 篇文章

订阅专栏

LeetCode

8 篇文章

订阅专栏

本文介绍了一种求解字符串中最长无重复字符子串长度的算法实现，通过使用集合和列表来跟踪当前子串的状态，确保子串中不包含重复字符，并记录最长子串的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

求一个字符串s里面最长的没有重复字符的子字符串的长度

有个问题，如果把list.size()直接放在if的判断条件里会出问题，原因未知，所以用了一个k

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int len = s.length();
        int maxx = 0;
        set<char>setc;
        list<char>listc;
        for (int i = 0; i < len; i++)
        {
            int numOfSet = setc.size();
            setc.insert(s[i]);
            int numOfSet2 = setc.size();
            listc.push_back(s[i]);
            if (numOfSet == numOfSet2) {
                while (listc.front() != s[i]) {
                    setc.erase(listc.front());
                    listc.pop_front();
                }
                listc.pop_front();
            }
            else {
                int k = listc.size();
                if (k > maxx) {
                    maxx = listc.size();
                }
            }
        }
        return maxx;
    }
};