LeetCode-Longest Substring Without Repeating Characters

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

简单来说就是维护一个没有重复字符的窗口，确定起点后先向后拓展终点，直到重复再向后移动起点直到无重复。

class Solution {
public:
    bool check(int*a,int n){
        for(int i=0;i<n;i++){
            if(a[i]>1)return false;
        }
        return true;
    }
    int lengthOfLongestSubstring(string s) 
    {
        if(s.length()<=0)return 0;
        int* all=new int[26];
        for(int i=0;i<26;i++)all[i]=0;
        int maxLength=1;
        int start=0;
        int end=0;
        all[s[start]-'a']++;
        
        while(true)
        {
            if(check(all,26))
            {
                if(end-start+1>maxLength)
                {
                    maxLength=end-start+1;
                }   
                end++;
                if(end>=s.length())break;
                all[s[end]-'a']++;
            }
            else
            {
                all[s[start]-'a']--;
                start++;
            }
            
        }
        delete all;
        return maxLength;
    }
};

public class Solution {
  public int lengthOfLongestSubstring(String s) {
        // Note: The Solution object is instantiated only once and is reused by
        // each test case.
        if (s.length() == 0)
            return 0;
        boolean[] count = new boolean[256];
        Arrays.fill(count, false);
        int start = 0, end = 0;
        int maxLen = 0;
        boolean flag = true;
        count[(int) s.charAt(0)] = true;
        end++;
        while (true) {
            if (end < s.length()) {
                if (count[(int) s.charAt(end)]) {
                    if (end - start > maxLen)
                        maxLen = end - start;
                    while (count[(int) s.charAt(end)]) {
                        count[(int) s.charAt(start)] = false;
                        start++;
                    }
                } else {
                    count[(int) s.charAt(end)] = true;
                    end++;
                }
            } else {
                if (end - start > maxLen)
                    maxLen = end - start;
                break;
            }
        }
        return maxLen;
    }
}