无重复最长子串

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s == null){
            return 0;
        }
        int l = 0;
        int r = 0;
        int res = 0;
        Set<Character> set = new HashSet<>();
        while(r< s.length()){
            char c = s.charAt(r);
            while(set.contains(c)){
                char d = s.charAt(l);
                set.remove(d); 
                l++;
            }
            set.add(c);
            res = Math.max(res, set.size());
            r++;
        }
        return res;
    }
}

字符串的所以字母异位词

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        Map<Character, Integer> windows = new HashMap<>();
         Map<Character, Integer> need = new HashMap<>();
         for(int i = 0; i < p.length(); i++){
             need.put(p.charAt(i), need.getOrDefault(p.charAt(i), 0)+1);
         }
         int left = 0;
         int right = 0;
         int vaildate = 0;
         List<Integer>  res = new ArrayList<>();
         while(right < s.length()){
             char c = s.charAt(right);
             right++;
             if(need.containsKey(c)){
                windows.put(c, windows.getOrDefault(c,0)+1);
                if(windows.get(c).equals(need.get(c))){
                    vaildate++;
                }
             }
             while(right -left >=  p.length()){
                 if(vaildate == need.size()){
                     res.add(left);
                 }
                 char d = s.charAt(left);
                 left++;
                 if(need.containsKey(d)){
                    if(windows.get(d).equals(need.get(d))){
                        vaildate--;
                    }
                    windows.put(d, windows.getOrDefault(d,0)-1);
             }
             }
         }
         return res;
    }
}