3. Longest Substring Without Repeating Characters(最长子串，双指针+hash)

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

暴力：时间0(n2) 

空间o(n)

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        const int n = s.size();
        std::unordered_set<char> s_set;
        int res = 0;
        for (int i = 0; i < n; i++) {
            int j = i;
            s_set.clear();
            for (; j < n ;++j) {
                if (s_set.find(s[j])==s_set.end()) {
                    s_set.emplace(s[j]);
                } else {
                    break;
                }
            }
            res = std::max(res,j-i);
        }
        return res;
    }
};

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        const int n = s.size();
        std::unordered_set<char> s_set;
        int res = 0;
        int j = 0;
        for (int i = 0; i < n;++i) {
            while(s_set.find(s[i])!=s_set.end()) {
                s_set.erase(s[j]);
                j++;
            } 
            res = std::max(res,i-j+1);
            s_set.emplace(s[i]);
        }   
        return res;
    }
};

这个题的关键在于重复的字符！

一个子串不可能出现重复的字符，那么我们可以用两个指针，一个指针用来遍历字符串，两个指针之间保持无重复字符，那么两个指针之间的长度即最大子串的长度。当发现有重复的字符时，另一个指针指向这个字符上一次出现的位置的下一个字符，继续遍历，直到找到最长子串。

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        j = 0
        maxlen = 0
        used ={}
        for i ,c in enumerate(s):
            if c not in used or  used[c] < j:
                 maxlen = max(maxlen,i-j + 1)
            else:
                j = used[c] + 1
            used[c] = i
        return maxlen

 代码中：

 used[c] < start  是为了处理"tmmzuxt" 这种连续出现的2个字母。