3 longest Substring without repeating characters (longest non-repeating contiguous substring medium)

In the string, the longest continuous substring is not repeated.

Idea: Use HashMap, find unordered_map faster than map, set a starting position, calculate the length, to reduce the value of the starting position

Eg:a,b,c,d,e,c,b,a,e

0 1 2) 3 4

0 1 5) 3 4

0 6 5) 3 4

7 6 5) 3 4

7 6 5 ) 3 8

When C appears again, the start value is set to Map[c]+1=3, and for the next B, because Map[b]<start, the direct map[b]=i

1 classSolution {2  Public:3     intLengthoflongestsubstring (strings) {4unordered_map<Char,int>Mymap;5         intflag=0;6         intstart=0, i=0;7unordered_map<Char,int>:: iterator ite;8          while(i<s.size ()) {9Ite=Mymap.find (S[i]);Ten             if(ite==Mymap.end ()) { Onemymap[s[i]]=i; AFlag=max (flag,i-start+1); -             } -             Else{ the                 if(mymap[s[i]]>=start) -start=mymap[s[i]]+1; -mymap[s[i]]=i; -Flag=max (flag,i-start+1); +             } -++i; +         } A         returnFlag; at     } -};

Time complexity: O (N)

Time-Out Solution:

1 classSolution {2  Public:3     intLengthoflongestsubstring (strings) {4map<Char,int>Mymap;5         intI=0, max=0, Flag;6map<Char,int>:: iterator ite;7          while(i<s.length ()) {8Ite=Mymap.find (S[i]);9             if(ite==Mymap.end ()) {Tenmymap[s[i]]=i; One                 if(Mymap.size () >max) Amax=mymap.size (); -             } -             Else{ theFlag=ite->second; - mymap.clear (); Every time it clears, it's the way it fails. -                  for(intj=flag+1; j<=i;++j) { -mymap[s[j]]=J; +                 } -             } +++i; A         } at         returnMax; -     } -};

3 longest Substring without repeating characters (longest non-repeating contiguous substring medium)