Longest Substring Without Repeating Characters

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

For example：

1.The longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3.

2.For "bbbbb" the longest substring is "b", with the length of 1.

算法主要用一个映射表记录现在已遍历到的字符，并将相应的索引记录下来。用一个起始索引变量记录字符串的开始索引位置，遇到重复字符（且该重复字符的索引值>=开始索引）时更新该索引，将其变为重复变量索引的后一位，并更新最大字符串长度。最后别忘了从开始索引到最后还没出现重复字符的情况。

startIndex：不包含重复字符的字符串开始索引，起始为0

maxLength：用于记录最长的非重复字符的字符串长度，起始为0

对于例1：主要信息随着遍历的改变情况如下

字符串

a

b

c

a

b

c

b

b

索引

0

1

2

3

4

5

6

7

a

0

0

0

3

3

3

3

3

b

 

1

1

1

4

4

6

7

c

 

 

2

2

2

5

5

5

startIndex

0

0

0

1

2

3

5

7

maxLength

0

0

0

3

3

3

3

3

class Solution {public:int lengthOfLongestSubstring(string s) {int maxLenght = 0;           //最长非重复字串长度int startIndex = 0;          //每次出现重复时更新开始索引map<char, int> record;       //用于记录字符出现索引的映射表for(int i=0;i<s.size();++i){if (record.count(s[i])>0&&record[s[i]]>=startIndex) //存在字符记录且索引不比开始索引小{maxLenght = max(maxLenght,i-startIndex);startIndex =record[s[i]]+1;   //出现重复字符，将开始索引更新为重复字符的后一位}record[s[i]] = i;}maxLenght = max(maxLenght,(int)s.size()-startIndex);return maxLenght;}};