LeetCode 003. Longest Substring Without Repeating

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.

看錯題了，題目要求找出字符串中的一個最長的子串，使得這個子串不包含重複的字符，這個子串的長度。當作了找出一個最長重複子串，它不包含重複的字符。題目給的兩個例子也剛好符合個人臆想，想一想也是醉了。

寫了一些臆想的代碼：

unordered_set<char> tmp;
bool is_unique(string& s)
{
    tmp.clear();
    for(int i=0; i!= s.size(); ++i)
    {
        if(tmp.find(s[i]) != tmp.end())
            return false;
        tmp.insert(s[i]);
    }
    return true;
}

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        unordered_set<string> allsubstrings;
        int maxlength = 1;
        int i, j;
        string str;
        for(i=0; i!= s.size(); ++i)
        {
            for(j =i+maxlength; j<= s.size(); ++j)
            {
                str = s.substr(i,j-i);
                if(is_unique(str))
                {
                    if(allsubstrings.find(str) != allsubstrings.end())
                    {
                        if(maxlength < str.size())
                            maxlength = str.size();
                    }
                    else
                    {
                        allsubstrings.insert(str);
                    }
                }
                else
                {
                    break;
                }
            }
        }
        return maxlength;
    }
};

原題的解：

int lengthOfLongestSubstring(string s) {
  int n = s.length();
  int i = 0, j = 0;
  int maxLen = 0;
  bool exist[256] = { false };
  while (j < n) {
    if (exist[s[j]]) {
      maxLen = max(maxLen, j-i);
      while (s[i] != s[j]) {
        exist[s[i]] = false;
        i++;
      }
      i++;
      j++;
    } else {
      exist[s[j]] = true;
      j++;
    }
  }
  maxLen = max(maxLen, n-i);
  return maxLen;
}