• 關(guān)鍵字：最長不重復(fù)子串、雙指針
• 難度：Medium
• 題目大意：求一個字符串最長不重復(fù)子串的長度

題目：

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：

使用兩個指針l、r分別記錄不重復(fù)子串的起始和終止位置，l、r初始值為0，使用set來放當(dāng)前不重復(fù)的字符，r向右移動：

• 字符不重復(fù)時，直接將字符add到set，r繼續(xù)向右移動；
• 字符重復(fù)時，l向右移動，同時set刪除掉字符串l位置的字符；
• 重復(fù)上述步驟；

思路實現(xiàn)：

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length()==0||s==null) return 0;
        int maxLen = 1;
        int l = 0;
        int r = 1;
        Set<Character> set = new HashSet<>();
        set.add(s.charAt(l));
        while(l<s.length()&&r<s.length()) {
            if(!set.contains(s.charAt(r))) {
                set.add(s.charAt(r));
                maxLen = Math.max(maxLen, r-l+1);
                r++;
            }
            else {
                set.remove(s.charAt(l));
                l++;
            }
        }
        return maxLen;
        
    }
}