Sliding window_

1. Power button _ The longest substring without repeating characters

/**
 *  Longest substring without repeating characters 
 *  Characters stored in set aggregate , If there is a repetition, directly from the previous set Collection removal , Add new characters 
 */

import java.util.HashSet;
import java.util.Set;

/**
 *  Ideas ： The sliding window 
 *  With (a)bcabcbb  The longest string to start with is  (abc)abcbb;
 *  With a(b)cabcbb  The longest string to start with is  a(bca)bcbb;
 *  With ab(c)abcbb  The longest string to start with is  ab(cab)cbb;
 *  With abc(a)bcbb  The longest string to start with is  abc(abc)bb;
 *  With abca(b)cbb  The longest string to start with is  abca(bc)bb;
 *  With abcab(c)bb  The longest string to start with is  abcab(cb)b;
 *  With abcabc(b)b  The longest string to start with is  abcabc(b)b;
 *  With abcabcb(b)  The longest string to start with is  abcabcb(b).
 *  Now observe , We need to remove characters while adding characters . Store characters one by one set aggregate , If the following element has repeated characters , be set The set element is out of the set until there is no such repeating element , Then put this element into set aggregate 
 */
public class Num48_ Longest substring without repeating characters  {
    public int lengthOfLongestSubstring(String s) {
        int max=0;
        int right=0;
        Set<Character> set = new HashSet<>();
        for (int left = 0; left < s.length(); left++) {
            // The left pointer traverses backward , remove left The previous element of 
            if(left!=0){
                set.remove(s.charAt(left-1));
            }
            // The left pointer is fixed , The right pointer moves backwards , Elements are not only put in set aggregate , Until I met with set The middle element repeats or goes to the empty record length 
            while (right<s.length()&&(!set.contains(s.charAt(right)))){
                set.add(s.charAt(right));
                right++;
            }
            max=Math.max(max,right-left);
        }
        return max;
    }
}