【leetcode】Longest Substring Without Repeating Characters

Longest Substring Without Repeating Characters

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.

举个例子：

abcaddbcef

abca遇到a后，我们下次从b开始比较

bcadd遇到d后，我们下次从d开始比较

dbcef结束

所以设置一个start_index作为起始位置，遇到重复的元素，起始位置设置在重复元素第一次出现的位置+1

 1 class Solution {
 2 public:
 3     int lengthOfLongestSubstring(string s) {
 4        
 5         map<char,int> char2index;
 6        
 7         int start_index=0;
 8         int result=0;
 9         int i;
10         for(i=0;i<s.length();i++)
11         {
12             //从来没有出现过的元素
13             if(char2index.find(s[i])==char2index.end())
14             {
15                 char2index[s[i]]=i;
16             }
17             else
18             {
19                 //出现过，且在start_index之后出现
20                 if(char2index[s[i]]>=start_index)
21                 {
22                     result=max(result,i-start_index);
23                     start_index=char2index[s[i]]+1;
24                     char2index[s[i]]=i;
25                 }
26                 else
27                 {
28                     //在start_index之前出现
29                     char2index[s[i]]=i;
30                 }
31  
32             }
33         }
34         result=max(result,i-start_index);
35        
36         return result;
37     }
38 };

简化版本：

 1 class Solution {
 2 public:
 3     int lengthOfLongestSubstring(string s) {
 4        
 5         map<char,int> char2index;
 6        
 7         int start_index=0;
 8         int result=0;
 9         int i;
10         for(i=0;i<s.length();i++)
11         {
12             if(char2index.find(s[i])!=char2index.end()&&char2index[s[i]]>=start_index)
13             {
14                 result=max(result,i-start_index);
15                 start_index=char2index[s[i]]+1;
16             }
17             char2index[s[i]]=i;
18         }
19        
20         result=max(result,i-start_index);
21        
22         return result;
23     }
24 };