[LeetCode] Longest Substring Without Repeating Characters 解题报告

寻找最长无重复字符子串

最新推荐文章于 2025-05-18 01:04:12 发布

原创最新推荐文章于 2025-05-18 01:04:12 发布 · 150 阅读

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本文介绍如何解决寻找字符串中最长无重复字符子串的问题，通过从左到右扫描并记录字符出现的位置来实现。提供了两种解决方案，包括初始化数组的方法、避免逻辑错误和确保捕获最后一个字符串的情况。

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.

» Solve this problem

[解题思路]
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。

[Code]
Version 1

1:    int lengthOfLongestSubstring(string s) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      int count[26];  
5:      if(s.size() ==0) return 0;  
6:      memset(count,-1, sizeof(count));  
7:      int start = 0;  
8:      int maxV = 0;  
9:      for(int i=0; i< s.size(); i++)  
10:      {  
11:        int index = s[i] - 97;  
12:        if(count[index] >= 0)  
13:        {            
14:          if(maxV < (i -start))  
15:          {  
16:            maxV = i-start;  
17:          }  
18:          i = count[index];  
19:          start = i+1;  
20:          memset(count,-1, sizeof(count));  
21:          continue;  
22:        }  
23:        count[index] = i;  
24:      }  
25:      if(maxV < (s.size() -start))  
26:      {  
27:        maxV = s.size()-start;  
28:      }  
29:      return maxV;  
30:    }

Version 2, refactor the code at 3/4/2013

1:       int lengthOfLongestSubstring(string s) {  
2:            int count[26];  
3:            memset(count, -1, 26*sizeof(int));  
4:            int len=0, maxL = 0;  
5:            for(int i =0; i< s.size(); i++,len++)  
6:            {  
7:                 if(count[s[i]-'a']>=0)  
8:                 {  
9:                      maxL = max(len, maxL);  
10:                      len =0;          
11:                      i = count[s[i]-'a']+1;  
12:                      memset(count, -1, 26*sizeof(int));   
13:                 }  
14:                 count[s[i]-'a']=i;  
15:            }      
16:            return max(len, maxL);  
17:       }

Note:
1. Line 3, since we store the index in array, so initializing the array as 0 will mistake the logic.
2. Line 16, catch the last string. for example, “abcd”, if no Line 16, it will just return 0 since the Line 9 won’t be triggered.