Longest Substring Without Repeating Character

LeetCode第三题，题目描述：

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.

错误的想法：

创建一个子串，遍历字符串的同时，首先判断当前字符是否在子串中，如果不在，则将其放到子串中；否则，清空子串，同时比较tmpLen与maxLen的大小，如果maxLen小于tmpLen，则用tmpLen替换maxLen，代码如下：

class Solution {public:    int lengthOfLongestSubstring(string s) {        int maxLength = 0, tmpLength = 0;    int startIndexOfMaxLength = 0, endIndexOfMaxLength = 0;    int i;    string currentSubStr;    int len = s.length();        for (i=0; i < len; ++i)    {    if (string::npos == currentSubStr.find(s[i]))//如果不存在    {    currentSubStr.append(1, s[i]);    tmpLength ++;    }    else    {    if (maxLength < tmpLength)    {    endIndexOfMaxLength = i;    startIndexOfMaxLength = i - currentSubStr.length();    maxLength = tmpLength;    tmpLength = 1;    currentSubStr = currentSubStr.erase(0, currentSubStr.length());    currentSubStr.append(1, s[i]);    }    }    }        if (maxLength < tmpLength)    maxLength = tmpLength;        return maxLength;    }};

这种做法是错误的，因为忽略这了这样一种情况：

Input: "dvdf"Output: 2Expected: 3

目前思路是：

在当前已知的最大长度的子串中，查找到第一个重复的字符，然后再从该字符处查找，直到查找到第二个重复