Longest Substring Without Repeating Characters

package cn.edu.xidian.sselab.hashtable;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

/**
*
* @author zhiyong wang
* title: Longest Substring Without Repeating Characters
* content:
*    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
*
*/
public class LongestSubString {

   //这种方法，时间超时，如果字符串非常长，时间复杂度为O(n^2)，肯定会超时，因为这是一个最优化问题，可以用动态规划来解
   public int lengthOfLongestSubString(String s){
       int length = s.length();
       if(length == 0) return 0;
       int max = 0;
       for(int i=0;i<length;i++){
           Set set = new HashSet();
           for(int j=i;j<length;j++){
               if(!set.add(s.charAt(j)) && set.size() > max){
                   max = set.size();
                   break;
               }
           }
       }
       System.out.println(max);
       return max;
   }
   //参考大牛的思路，用一个map来保存，key表示s中的字符，value表示字符的下标，
   //同时用两个指针，第一个指向字符的下标，第二个指向出现相同的字符的下一个字符
   //时间复杂度为O(n)
   public int lengthOfLongestSubStrings(String s){
       int length = s.length();
       if(length == 0) return 0;
       int max = 0;
       Map<Character,Integer> map = new HashMap();
       for(int i=0,j=0;i<length;i++){
           if(map.containsKey(s.charAt(i))){
               j = Math.max(j, map.get(s.charAt(i))+1);
           }
           map.put(s.charAt(i), i);
           max = Math.max(max, i - j + 1);
       }
       return max;
   }
}

来源：https://www.cnblogs.com/wzyxidian/p/5167748.html

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substring