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.

• Set java类型 TreeSet HashSet

暴力法

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int ans = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j <= s.length(); j++) {
                if (isUnique(s, i, j)) {
                    ans = Math.max(j - i, ans);
                }
            }
        }
        return ans;
    }
    public boolean isUnique(String s, int i, int j) {
        Set<Character> chSets = new HashSet<>();
        for (int z = i; z < j; z++) {
            Character ch = s.charAt(z);
            if (chSets.contains(ch)) return false;
            chSets.add(ch);
        }
        return true;
    }
}

时间复杂度O(n3)

划窗口法

public int lengthOfLongestSubstring(String s) {
        int i = 0, j = 0;
        int n = s.length();
        int ans = 0;
        Set<Character> chSets = new HashSet<>();
        while (i < n && j < n) {
            if (!chSets.contains(s.charAt(j))) {
                chSets.add(s.charAt(j++));
                ans = Math.max(j-i, ans);
            }
            else {
                chSets.remove(s.charAt(i++));
            }
        }
        
        return ans;
    }

时间复杂度： O（2n）
空间复杂度： O（min(m, n)）

opt sliding

public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        int ans = 0;
        Map<Character, Integer> chMaps = new HashMap<>();
        for (int i = 0, j = 0; j < n; j++) {
            if (chMaps.containsKey(s.charAt(j))) {
                i = Math.max(i, chMaps.get(s.charAt(j)));
            }
            ans = Math.max(ans, j - i + 1);
            chMaps.put(s.charAt(j), j + 1);
        }
        return ans;
    }

滑窗口时 i 也会遍历n， 遇到想重复的char ， i 直接跳到 临近重复的下一个，避免遍历。