Leetcode: Longest Substring Without Repeating Characters

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.

Use HashSet

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

Use HashMap

   public int lengthOfLongestSubstring(String s) {
        if (s.length()==0) return 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int max=0;
        for (int i=0, j=0; i<s.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;
    }

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