"Leetcode" "longest Substring without repeating characters" "the oldest string of non-repeating characters" "C + +"

• Title: Given a string, the length of the oldest string with no repeating character is obtained.
• Idea: For loop once, time complexity is O (N). The ASCII value of the character is 32~126. Start represents the initial position of the current non-repeating character string, with an initial value of 0, and a positional array of pos[128] that represents the position of the corresponding character at the current position of the for Loop index. If the current character is S[i] (its ASCII value is Cur_pos), if Pos[cur_pos]>=start, indicating that the character S[i] after start, the substring starting with start encounters a repeating character for the first time, stop. Determine whether the current length is greater than Max_len, or update max_len if it is greater. Then update start, and Pos[cur_pos] .... Note the processing of the last character.
• Code:

  classSolution { Public:    intLengthoflongestsubstring (strings) {intlen_s=s.size (); Const intm= -; intPos[m];  for(intI=0; i<m;i++) Pos[i]=-1; intstart=0; intmax_len=0;  for(intI=0; i<len_s;i++)        {            intcur_pos=S[i]; if(pos[cur_pos]>=start) {                if(max_len< (istart)) {Max_len=i-start; } Start=pos[cur_pos]+1; } Pos[cur_pos]=i; }        if((Len_s-start) >Max_len) Max_len=len_s-start; returnMax_len; }};

"Leetcode" "longest Substring without repeating characters" "the oldest string of non-repeating characters" "C + +"