Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given “abcabcbb”, the answer is “abc”, which the length is 3.

Given “bbbbb”, the answer is “b”, with the length of 1.

Given “pwwkew”, the answer is “wke”, with the length of 3. Note that the answer must be a substring, “pwke” is a subsequence and not a substring.

注释：给定一个字符串，求出该字符串最长的无重复子串的长度。

以”abcabcbb”为例进行分析，预先设定一个队列queue和字符串数组exist，队列当前长度cur，最大长度max：
‘a’ : exist中不包含’a’，加入exist和queue中，cur = 1, max = 1;
‘b’ : exist中不包含’b’, 加入exist和queue中，cur = 2, max = 2；
‘c’ : exist中不包含’c’, 加入exist和queue中， cur = 3 , max = 3;
此时队列queue : {‘a’, ‘b’, ‘c’}
注：当新增某个字符时，需要判断cur和max，若cur大于max，更新max值；当移除字符时，cur一定小于等于max，不需要更新

‘a’ : exist中包含’a’, 为判断后面连续字符串长度，将queue中’a’移除，将这个’a’加入队尾;cur =3 ; max = 3; queue={‘b’,’c’,’a’}
‘b’ : exist中包含’b’, 为判断后面连续字符串长度，将queue中’b’移除，将这个’b’加入队尾;cur =3 ; max = 3; queue={‘c’,’a’,’b’}
‘c’ : exist中包含’c’, 为判断后面连续字符串长度，将queue中’c’移除，将这个’c’加入队尾;cur =3 ; max = 3; queue={‘a’,’b’,’c’}

‘b’: exist中包含’b’, 为判断后面连续字符串长度，将queue中‘b’及之前的元素移除，将这个’b’加入队尾;cur =2 ; max = 3; queue={‘c’,’b’}
‘b’: exist中包含’b’, 为判断后面连续字符串长度，将queue中‘b’及之前的元素移除，将这个’b’加入队尾;cur =1 ; max = 3; queue={‘b’}

实现代码：

public int lengthOfLongestSubstring(String s) {   int len = s.length();    Queue<Character> queue = new LinkedList<Character>();    List<Character> exist = new ArrayList<Character>();    int max = 0 ;    int cur = 0 ;    for(int i = 0; i < len; i++) {        char ch = s.charAt(i);        if(!exist.contains(ch)){            exist.add(ch);            queue.offer(ch);            cur += 1;            if(cur > max) {                max = cur;            }        } else {            while(queue.peek() != ch) {                char curCh = queue.poll();                exist.remove((Character)curCh);                cur--;            }            queue.poll();            queue.offer(ch);        }    }    return max; }