Solution URL
https://leetcode.com/submissions/detail/412071922/
代码
class Solution {
public String decodeString(String s) {
//ans: https://leetcode-cn.com/problems/decode-string/solution/decode-string-fu-zhu-zhan-fa-di-gui-fa-by-jyd/
//Time: O(N) Space: O(N)
StringBuilder res = new StringBuilder();
int multi = 0;
LinkedList<Integer> stack_multi = new LinkedList<>();
LinkedList<String> stack_res = new LinkedList<>();
//遍历s中所有字符（包括数字，字母，左括号，右括号）
for(Character c: s.toCharArray()){
//如果是左括号 [
if(c.equals('[')){
//将数字和字符分别入栈，并且还原multi和res
stack_multi.add(multi);
stack_res.add(res.toString());
multi = 0;
res = new StringBuilder();
}
//如果是右括号 ]
else if(c.equals(']')){
StringBuilder temp = new StringBuilder();
int cur_multi = stack_multi.removeLast();//弹出一次数字
//根据弹出的数字n将之前的res，复制n次
for(int i = 0; i < cur_multi; i++) temp.append(res);
//将存字母的stack中元素弹出，和复制完的字符组合起来
res = new StringBuilder(stack_res.removeLast() + temp);
}
//如果是数字,则赋值给multi(当然要处理一下比如12这样的数字)
else if(c >= '0' && c <= '9') multi = multi * 10 + Integer.parseInt(c + "");
//如果是字母
else res.append(c);
}
return res.toString();
}
}
