一、 问题描述
Leecode第三题，题目为：
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: “abcabcbb”
Output: 3
Explanation: The answer is “abc”, with the length of 3.
Example 2:
Input: “bbbbb”
Output: 1
Explanation: The answer is “b”, with the length of 1.
Example 3:
Input: “pwwkew”
Output: 3
Explanation: 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.
先把问题翻译一下：
给定一个字符串，在字符不重复的情况下返回最长子字符串的长度。
二、解题思路
（1）从字符串第一个字符至最后一个字符顺序遍历字符串，并标记该不重复字符串的长度，for example：len=k,(k为不重复字符串的长度)；
（2）继续遍历字符串，如果有不重复字符串length大于已标记字符串长度，那么更新符号变量的值，否则，继续遍历，直至最后一个字符串。
（3）返回最长字符串的内容及长度。
三、实现代码
class Solution {
public:int lengthOfLongestSubstring(string s) {string max_string = ""; string goal = ""; char ch; for(int i=0; i<s.length(); i++){ ch = s[i];int idx = goal.find(ch);if(idx==string::npos){ goal += s[i];} else{ goal = goal.substr(idx+1);goal += s[i];} if(goal.length()>=max_string.length()){ max_string = goal;} } return max_string.length();}
};