3. 无重复字符的最长子串

class Solution {
  int DupFromEnd(string &S, int Begin, int End) {
    char Back = S[End - 1];
    for (int i = End - 2; i >= Begin; --i) {
      if (S[i] == Back)
        return i;
    }
    return -1;
  }

public:
  int lengthOfLongestSubstring(string s) {
    int Longest = 0;
    int Left = 0, Right = 0;
    while (Right < s.length()) {
      // Fetch
      int DupIdx = DupFromEnd(s, Left, Right + 1);

      // Move
      if (DupIdx == -1)
        ++Right;
      else
        Left = DupIdx + 1;

      Longest = std::max(Longest, Right - Left);
    }
    return Longest;
  }
};

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        // 哈希集合，记录每个字符是否出现过
        unordered_set<char> occ;
        int n = s.size();
        // 右指针，初始值为 -1，相当于我们在字符串的左边界的左侧，还没有开始移动
        int rk = -1, ans = 0;
        // 枚举左指针的位置，初始值隐性地表示为 -1
        for (int i = 0; i < n; ++i) {
            if (i != 0) {
                // 左指针向右移动一格，移除一个字符
                occ.erase(s[i - 1]);
            }
            while (rk + 1 < n && !occ.count(s[rk + 1])) {
                // 不断地移动右指针
                occ.insert(s[rk + 1]);
                ++rk;
            }
            // 第 i 到 rk 个字符是一个极长的无重复字符子串
            ans = max(ans, rk - i + 1);
        }
        return ans;
    }
};

class Solution {
public:
  int lengthOfLongestSubstring(string s) {
    int Longest = 0;
    int Begin = 0;
    unordered_map<char, int> CharIdx;
    for (int i = 0; i < s.size(); ++i) {
      if (auto I = CharIdx.find(s[i]); I != CharIdx.end())
        Begin = I->second + 1;
      CharIdx[s[i]] = i;
      Longest = max(Longest, i - Begin + 1);
    }
    return Longest;
  }
};

class Solution {
public:
  int lengthOfLongestSubstring(string s) {
    int Longest = 0;
    int l = 0;
    unordered_set<char> Charset;
    for (int i = 0; i < s.size(); ++i) {
	  // If charset overlaps, erase from the left ONE BY ONE.
      while (Charset.count(s[i]))
        Charset.erase(s[l++]);
      Charset.insert(s[i]);
      Longest = max(Longest, i - l + 1);
    }
    return Longest;
  }
};

//外层循环扩展右边界，内层循环扩展左边界
for (int l = 0, r = 0 ; r < n ; r++) {
	//当前考虑的元素
	while (l <= r && check()) {//区间[left,right]不符合题意
        //扩展左边界
    }
    //区间[left,right]符合题意，统计相关信息
}

class Solution {
public:
  int lengthOfLongestSubstring(string s) {
    int Result = 0;
    int Left = 0;
    unordered_set<char> Set;
    for (int i = 0; i < s.length(); ++i) {
      while (Set.contains(s[i]))
        Set.erase(s[Left++]);
      Set.insert(s[i]);
      Result = max(Result, i - Left + 1);
    }
    return Result;
  }
};