• 使用栈的特性，遇到 】 符号从栈里面取出 ```java package com.example.suanfa.alg1;

    import java.util.LinkedList;

    /**

    • @author lfy

    • @date 2022-05-25 */ public class DecodeString {

      public String decodeString(String s) {

      1.  StringBuilder res = new StringBuilder();
      2.  int multi = 0;
      3.  LinkedList<Integer> stack_multi = new LinkedList<>();
      4.  LinkedList<String> stack_res = new LinkedList<>();
      5.  for(Character c : s.toCharArray()) {
      6.      if(c == '[') {
      7.          stack_multi.addLast(multi);
      8.          stack_res.addLast(res.toString());
      9.          multi = 0;
      10.          res = new StringBuilder();
      11.      }
      12.      else if(c == ']') {
      13.          StringBuilder tmp = new StringBuilder();
      14.          int cur_multi = stack_multi.removeLast();
      15.          for(int i = 0; i < cur_multi; i++){
      16.              tmp.append(res);
      17.          }
      18.          res = new StringBuilder(stack_res.removeLast() + tmp);
      19.      }
      20.      else if(c >= '0' && c <= '9') multi = multi * 10 + Integer.parseInt(c + "");
      21.      else res.append(c);
      22.  }
      23.  return res.toString();

      }

      public static void main(String[] args) {

       String s = "3[a]2[bc]";
 String s1 = "3[a2[c]]";
 DecodeString test = new DecodeString();
 System.out.println(test.decodeString(s1));

      } } ```

    • 递归法调用 ```java class Solution { public String decodeString(String s) {
        return dfs(s, 0)[0];
      } private String[] dfs(String s, int i) {
        StringBuilder res = new StringBuilder();
  int multi = 0;
  while(i < s.length()) {
      if(s.charAt(i) >= '0' && s.charAt(i) <= '9') 
          multi = multi * 10 + Integer.parseInt(String.valueOf(s.charAt(i))); 
      else if(s.charAt(i) == '[') {
          String[] tmp = dfs(s, i + 1);
          i = Integer.parseInt(tmp[0]);
          while(multi > 0) {
              res.append(tmp[1]);
              multi--;
          }
      }
      else if(s.charAt(i) == ']') 
          return new String[] { String.valueOf(i), res.toString() };
      else 
          res.append(String.valueOf(s.charAt(i)));
      i++;
  }
  return new String[] { res.toString() };
      } 
    public static void main(String[] args) {
    String s = "3[a]2[bc]";
    String s1 = "3[a2[c]]";
    DecodeString test = new DecodeString();
    System.out.println(test.decodeString(s1));
}

    }

    ```