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.

---

给定一个字符串，求出它最长的没有重复字符的子串的长度。
法一：定义字符串str、rst，遍历给定的字符串s，每遍历一个字符c，判断它是否在字符串str中存在，若不存在，直接添加到字符串str末尾，若存在，比较字符串str与rst的长度，若前者比后者长，将字符串赋给rst，并抹去字符串str从0到字符c处的所有字符。当遍历结束时，字符串rst即为给定字符串s的最长不含重复字符的子串。当当前最长子串（即字符串rst）的长度比 剩余字符数+当前字符串长度（即字符串str） 还长时，没必要继续遍历下去，因为已经不可能存在比字符串rst还长的子串了。代码如下：

    int lengthOfLongestSubstring(std::string s) {
        size_t pos = 0;
        std::string str, rst;
        int cur = 0, total = s.length();
        for (auto c : s)
        {
            pos = str.find(c);
            if (pos != std::string::npos)
            {
                if (str.length() > rst.length())
                    rst = str;
                str = str.substr(pos + 1);
                if (rst.length() >= str.length() + total - cur)
                    break;
            }
            str.append(1, c);
            cur++;
        }
        int cur_len = str.length();
        int record_len = rst.length();
        return cur_len > record_len ? cur_len : record_len;
    }

 法二：法一中遍历给定字符串是无法避免的，查找字符str是否含有字符c可以用HashMap来做，因为最终只需要最长不含重复字符的子串的长度，我们甚至不需要保存临时字符串str，改进后的代码如下：

    int lengthOfLongestSubstring(std::string s) {
        std::unordered_set<char> set;
        int left = 0,length = 0, cnt = 0, cur = 0, total = s.length();
        for (char c : s)
        {
            if (set.find(c) != set.end())
            {
                cnt = set.size();
                if (cnt > length)
                    length = cnt;
                char ch = '\0';
                while (set.find(c) != set.end())
                {
                    ch = s.at(left);
                    auto iter = set.find(ch);
                    if (iter != set.end())
                        set.erase(iter);
                    left++;
                }
            }
            set.insert(c);
            cur++;
            if (length >= (int)set.size() + total - cur)
                break;
        }
        cnt = set.size();
        return length > cnt ? length : cnt;
    }