1 class Solution {
2 public int ptr = 0;
3 public String decodeString(String s) {
4 LinkedList<String> stk = new LinkedList<>();
5 while(ptr < s.length()){
6 char cur = s.charAt(ptr);
7 System.out.println(ptr+" "+cur);
8 if(Character.isDigit(cur)){
9 stk.addLast(getDigit(s));//【可能是多位的，一个完整的数，不一定是单个数字】
10 }
11 else if(Character.isLetter(cur) || cur == '['){
12 stk.addLast(String.valueOf(s.charAt(ptr)));
13 ptr++;
14 }
15 else{
16 ++ptr;
17 LinkedList<String> sub = new LinkedList<>();
18 while(!"[".equals(stk.peekLast())){
19 sub.addLast(stk.removeLast());
20 }
21 stk.removeLast();
22 Collections.reverse(sub);
23 int repeatNum = Integer.valueOf(stk.removeLast());
24 String curStr = getString(sub);
25 StringBuilder sb = new StringBuilder();
26 while(repeatNum > 0){
27 sb.append(curStr);
28 repeatNum--;
29 }
30 stk.addLast(sb.toString());
31 }
32 }
33 return getString(stk);
34 }
35 public String getString(LinkedList<String> sub){
36 StringBuilder sb = new StringBuilder();
37 for(String s: sub){
38 sb.append(s);
39 }
40 return sb.toString();
41 }
42 public String getDigit(String s){
43 StringBuilder sb = new StringBuilder();
44 while(Character.isDigit(s.charAt(ptr))){
45 sb.append(s.charAt(ptr));
46 ptr++;
47 }
48 return sb.toString();
49 }
50 }
