Longest Substring Without Repeating Characters

import java.util.Scanner;

/**
 * Created by qmq
 * Given a string, find the length of the longest substring without repeating characters.
 * Example 1:
 * Input: "abcabcbb"
 * Output: 3
 * Explanation: The answer is "abc", with the length of 3.
 *
 * @ 18-10-12
 **/
class Solution7 {
    public int longestSubstring(String s) {
        if (s.length() < 2) {
            return s.length();
        }
        int max = -1, start = 0;
        for (int i = 1; i < s.length(); i++) {
            if (i < s.length()) {
                int index = s.substring(start, i).indexOf(s.charAt(i));
                if (index != -1) {
                    max = Math.max(max, i - start);
                    start += (index + 1);
                } else {
                    max = Math.max(max, i - start);
                }
            }
        }
        return max != -1 ? max : s.length();
    }
}

public class LongestSubstring {
    public static void main(String[] args) {
        Solution7 sol = new Solution7();
        Scanner scanner = new Scanner(System.in);
        String str = scanner.nextLine();
        System.out.println(sol.longestSubstring(str));
    }
}

#include <bits/stdc++.h>
using namespace std;

// Given a string, find the length of the longest substring without repeating characters.
// Example 1 :
// Input : "abcabcbb" Output : 3 Explanation : The answer is "abc",
// with the length of 3.
// the first thought is to round the first situation charactor

// new hashmap
// hashmap.put(x)
// cnt(x)
// if(x>1)
// cnt the number[xi]
// new hashmap
// continue
// until find
// return the number[xi].max

class Solution
{
  public:
    int LongestSubstring(char *s)
    {
        int len = 0;
        char *end = s, *temp;
        char *addressTable[128] = {NULL};
        while (*end) //非空地址
        {
            temp = addressTable[*end]; //哈希表
            addressTable[*end] = end;  //哈希表end位置初始化
            if (temp >= s)             //当前子域大于历史域，则进行域长度更新
            {
                len = end - s > len ? end - s : len;
                s = temp + 1;
            }
            end++;
        }
        len = end - s > len ? end - s : len; //排除最后一次迭代影响
        return len;
    }
};
int main()
{
    char *str;
    cin >> str;
    Solution sol;
    cout << sol.LongestSubstring(str) << endl;
    return 0;
}