leetcode 3. Longest Substring Without Repeating Characters

题目描述：

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. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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.

思路1：滑动窗口的思想，slow,fast指针来模拟窗口

用HashSet来保存当前最长的不含重复字符的字符串，我们用快指针来判断当前字符是否存在于HashSet中，若不存在，则加入set， 若已经存在，则用slow指针从前开始一出，直到移除重复的字符后，再次加入新的字符。在遍历的过程中，不断的判断更新最长不重复字串的长度

实现1：

public int lengthOfLongestSubstring(String s) {
        int slow=0,fast=0,ret=0;
        HashSet set=new HashSet<>();
        while(fast

实现2：

和上面思路一样

 public int lengthOfLongestSubstring2(String s) {
        HashSet set=new HashSet<>();
        int ret=0;
        int tail=0;
        for(int head=0;head

思路2：

用哈希表存储，若存在重复的，则记下重复元素的索引，并在遍历过程中不断更新最大长度。

 public int lengthOfLongestSubstring3(String s) {
        int maxlen=0;
        int tail=-1;//用来记录与当前字符重复的字符所在的索引位置
        HashMap map=new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            Character ch=s.charAt(i);
            Integer old=map.get(ch);
            map.put(ch,i);
            if(old!=null&&old>tail){
                tail=old;
            }
            maxlen=Math.max(maxlen,i-tail);
        }
        return maxlen;
    }