Leetcode3. Longest Substring Without Repeating Characters (the longest substring without repeating characters)

The main idea of ​​the title: Find the longest substring of a given string, requiring no repetition of characters in the substring. The difference between a substring and a subsequence is that a substring must be composed of consecutive characters, while a subsequence allows characters to be interrupted.

Topic Analysis:

Use num[i] to represent the longest substring ending with the character s[i]. We only need to scan the pointer forward from i and stop counting if there are repeated characters. The obtained value is the longest substring ending with s[i]. The length of the string. The time complexity is O(n^2)

Code display:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int len = s.length();
        bool visit[256];
        memset(visit,0,sizeof(visit));
        int count = 0;
        int max = 0;
        for(int i=0;i<len;i++){
            for(int j=i;j>=0;j--){
                if(!visit[s[j]]){
                    count++;
                    visit[s[j]] = 1;
                }
                else
                    break;
            }
            if(count>max)
                max = count;
            memset(visit,0,sizeof(visit));
            count = 0;
        }
        return max;
    }
};

The second solution is that the longest substring is s[start,...,i], and the length of the substring can be returned. The point is that start and i change dynamically in one loop. For example, s="abcabcbb". 

First, start points to -1 position, i is 0 position, pos['a']=0, return length i-start=1;

Then i=1, there is no repetition, start does not change, pos['b']=1, return i-start=2;

Then i=2, there is no repetition, start does not change, pos['c']=2, return i-start=3;

Then i=3, there is a repetition, start is moved back, start = pos['a'] = 0, update pos['a'] = 4, return i-start=3  …

Why is start moved backwards here? Because the substring is required to be non-repetitive, if the start is not moved backward, it must be abca, and there is a repeated a. Therefore, once there is a repetition, it must be moved backward by start, and then moved to the position where a appeared for the first time.

Code display:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int maxlen = 0;
        int pos[256];
        memset(pos,-1,sizeof(pos));
        int start = -1;
        for(int i=0;i<s.length();i++){
            if(pos[s[i]]>start)
                start = pos[s[i]];
            pos[s[i]] = i;
            maxlen = max(maxlen,i-start);
        }
        return maxlen;
    }
};