Leetcode 387. First Unique Character in a String

Problem:given a string, find the first non-repeating character in it and return it ' s index. If it doesn ' t exist, return-1.

Example:

s = "Leetcode" return 0.s = "Loveleetcode", return 2.

The first idea is to open an alphabet counter, traverse the elements in string s, and count each letter with a counter. Finally, traversing s again, looking for a count of 1 characters and then returning its index. The solution code is as follows:

1 classSolution {2  Public:3     intFirstuniqchar (strings) {4         int*alphabet =NULL;5Alphabet =New int[ -]();6          for(inti =0; I < s.length (); i++)7         {8alphabet[int(S[i]-'a')]++;9         }Ten          for(inti =0; I < s.length (); i++) One         { A             if(alphabet[int(S[i]-'a')] ==1) -             { -                 returni; the             } -         } -         return-1; -          +          -     } +};

However, if the string is very long and two traversal is expensive, you can consider a traversal, record the number and index of each letter with a hash table, with the following code:

1 classSolution {2  Public:3     intFirstuniqchar (strings) {4unordered_map<Char, pair<int,int>>m;5         intIDX =s.length ();6          for(inti =0; I < s.length (); i++)7         {8m[s[i]].first++;9M[s[i]].second =i;Ten         } One          for(Auto &p:m) A         { -             if(P.second.first = =1) -             { theIDX =min (idx, p.second.second); -             } -         } -         returnIDX = = S.length ()? -1: idx; +     } -};

Leetcode 387. First Unique Character in a String