【Leetcode】394.字符串解码

class Solution {
    public int ptr = 0;
    public String decodeString(String s) {
        LinkedList<String> stk = new LinkedList<>();
        while(ptr < s.length()){
            char cur = s.charAt(ptr);
            System.out.println(ptr+" "+cur);
            if(Character.isDigit(cur)){
                stk.addLast(getDigit(s));//【可能是多位的，一个完整的数，不一定是单个数字】
            }
            else if(Character.isLetter(cur) || cur == '['){
                stk.addLast(String.valueOf(s.charAt(ptr)));
                ptr++;
            }
            else{
                ++ptr;
                LinkedList<String> sub = new LinkedList<>();
                while(!"[".equals(stk.peekLast())){
                    sub.addLast(stk.removeLast());
                }
                stk.removeLast();
                Collections.reverse(sub);
                int repeatNum = Integer.valueOf(stk.removeLast());
                String curStr = getString(sub);
                StringBuilder sb = new StringBuilder();
                while(repeatNum > 0){
                    sb.append(curStr);
                    repeatNum--;
                }
                stk.addLast(sb.toString());
            }
        }
        return getString(stk);
    }
    public String getString(LinkedList<String> sub){
        StringBuilder sb = new StringBuilder();
        for(String s: sub){
            sb.append(s);
        }
        return sb.toString();
    }
    public String getDigit(String s){
        StringBuilder sb = new StringBuilder();
        while(Character.isDigit(s.charAt(ptr))){
            sb.append(s.charAt(ptr));
            ptr++;
        }
        return sb.toString();
    }
}