改进了一下你的代码，虽然结果正确，但超时了 import java.util.*; import java.util.regex.*; public class Main{ public static void main(String[] args){ Scanner in = new Scanner(System.in); while(in.hasNext()){ String s1 = in.nextLine().toLowerCase(); String s2 = in.nextLine().toLowerCase(); System.out.println(helper(s1, s2, 0, 0)); } } private static boolean helper(String s1, String s2, int p1, int p2){ String pattern="[^0-9^a-z^A-Z]"; //base case if (p1 == s1.length() && p2 == s2.length()){ return true; }else if (p1 == s1.length() || p2 == s2.length()){ return false; } //遇到'*'两种情况，要不就各跳过一个比较后面，要不就s2继续往后跳先不比较 //加入正则表达式，匹配到不合法就false，以及添加匹配0个字符 if (s1.charAt(p1) == '*'){ boolean ismatch=Pattern.matches(pattern,s2.charAt(p2)+""); if(ismatch) return false; return helper(s1, s2, p1, p2+1) || helper(s1, s2, p1+1, p2+1)||helper(s1, s2, p1+1, p2); //遇到'?'跟两个一样操作一样，直接指针都往后移一个继续比较 }else if (s1.charAt(p1) == '?'){ boolean ismatch=Pattern.matches(pattern,s2.charAt(p2)+""); if(ismatch) return false; return helper(s1, s2, p1+1, p2+1); }else if(s1.charAt(p1) == s2.charAt(p2)){ return helper(s1, s2, p1+1, p2+1); }else { return false; } } }