Leetcode | | Longest Substring without repeating characters (O (n) algorithm) problem

Problem:

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "ABCABCBB" are "abc", which the length is 3. For "bbbbb" the longest substring are "B", with the length of 1.

Thinking:

(1) Looking for substrings, to do a traversal.

(2) Brute force hack method, check each substring, use hash table mapping idea, open an array, subscript the ASCII code for the character, because there is no description of the type of string, you can take some measures to compress the size of hash table. The worst time complexity of brute force is O (n*n)

(3) found an O (n) algorithm, great

In fact, much of the work here is repetitive and useless.
See an example:
S= "Abbdeca".
T1= "Abbdeca", t1[1]==t1[2].
T2= "Bbdeca", t2[0]==t2[1].
T3= "Bdeca", has been scanned to the last.
t4= "Deca", T5, T6, T7 all ibid.
We have scanned the s[2] while processing the T1, and then scanned s[2] to s[6 when processing T3, and the two substrings have scanned the entire string.
In other words, there are only two places where the substring can stop scanning: 1.s[2];2.s[6] (end).
For another example, s= "Aaab", the location where the substring can stop scanning is: s[1],s[2],s[3] (end).


So we can consider just scanning the parent string and removing the longest non-repeating substring directly from the parent string.
For S[i]:
1.s[i] does not appear in the current substring, then the length of the substring plus 1;
2.s[i] appears in the current substring, the position of the subscript is J, then the starting position of the new substring must be greater than J, in order to make the new substring as long as possible, so the starting position is selected as j+1.

Code

Description: The solution complexity of the annotation is O (n*n), the bit O (n) is not commented

#include <iostream> #include <string> #include <memory.h>using namespace Std;/*class solution {public        : int lengthoflongestsubstring (string s) {int a[100];//Compressed hash table memset (a,0,sizeof (int) *100);        int length = S.size ();        int index=0;        int max=0;            for (int i=0;i<length;i++) {int j=i;                while ((j<length) && (a[s.at (j) -32]==0)//Compression hash Table {a[s.at (j) -32]=1;                index++;            j + +;            } memset (a,0,sizeof (int) *100);            max= (Max>index) Max:index;            index=0;        if (j==length-1)//Here also has a small trick that can effectively reduce the time complexity of break;    } return max;        }};*/class Solution {Public:int lengthoflongestsubstring (string s) {//Start typing your C + + solution below        Do not write int main () function int locs[256];//holds the position of the last occurrence of the character memset (locs,-1, sizeof (locs)); int IDx =-1, max = 0;//idx is the start position of the current substring -1 for (int i = 0; i < s.size (); i++) {if (Locs[s[i]] > IDX            )//If the current character appears, the starting position of the current substring is the position of the last occurrence of the character +1 {idx = locs[s[i]];            } if (I-idx > Max)//Here is the key!!!!!!!!!            {max = I-idx;        } Locs[s[i]] = i;    } return max;    }};int Main () {string str = "Abcdab";    Solution MySolution; Cout<<mysolution.lengthoflongestsubstring (str) <<endl;}

Leetcode | | Longest Substring without repeating characters (O (n) algorithm) problem