394. 字符串解码

class Solution {
        public String decodeString(String s) {
            //括号前的数量，用于做倍数计算
            LinkedList<Integer> countStack=new LinkedList<>();
            //字符串栈
            LinkedList<String> strStack=new LinkedList<>();
            StringBuilder sb=new StringBuilder();

            int count=0;
            //eg：输入：s = "3[a]2[bc]"
            for (char c : s.toCharArray()) {
                if (c>='0'&&c<='9'){
                    count=count*10+(c-'0');
                }else if (c>='a'&&c<='z'){
                    sb.append(c);
                }else if (c=='['){
                    countStack.push(count);
                    strStack.push(sb.toString());

                    sb=new StringBuilder();
                    count=0;
                }else {
                    //c==]
                    StringBuilder currSb=new StringBuilder();
                    currSb.append(strStack.pop());
                    Integer currCount = countStack.pop();
                    for (int i = 0; i < currCount; i++) {
                        currSb.append(sb);
                    }
                    sb=currSb;
                }
            }

        return sb.toString();
        }
    }