直接O(n),我理解判断长度就可以，不需要再比较子串了 import java.util.*; public class Main {     public static void main(String[] args) {         Character[] arr = new Character[]{'a&(417)#39;,'b&#39;,'c&(2248)#39;,'d&#39;};         LinkedList<Character> word = new LinkedList<>(Arrays.asList(arr));         String s = "tbcacbdata";         LinkedList<Character> window = new LinkedList<>();         int index =1;         for(char c: s.toCharArray()){             // 如果是word中出现的字母             if(word.contains(c)){                 // 首先移动windows到重复字母之后                 if(window.contains(c)){                     while (window.pollFirst() != c);                 }                 window.add(c);                 // 判断是否存在答案                 if(window.size()==word.size()){                     System.out.println(index-word.size());                     return;                 }             }             // 不是 word中的字母             else {                 window.clear();             }             index++;         }     } }