Given an encoded string, return it's decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

一刷
题解：
利用stack求解，思路同385

public class Solution {
    public static String decodeString(String s) {
        String res = "";
        Stack<Integer> countStack = new Stack<>();
        Stack<String> resStack = new Stack<>();
        int idx = 0;
        while(idx<s.length()){
            if(Character.isDigit(s.charAt(idx))){
                int count = 0;
                while(Character.isDigit(s.charAt(idx))){
                    count = 10*count + (s.charAt(idx) - '0');
                    idx++;
                }
                countStack.push(count);
            }
            else if(s.charAt(idx) == '['){
                resStack.push(res);
                res = "";
                idx++;
            }
            else if(s.charAt(idx) == ']'){
                StringBuilder temp = new StringBuilder(resStack.pop());
                int repeatTimes = countStack.pop();
                while(repeatTimes>0){
                    temp.append(res);
                    repeatTimes--;
                }
                res = temp.toString();
                idx++;
            }
            else{
                res += s.charAt(idx++);
            }
        }
        return res;
    }
}

二刷

public class Solution {
    public String decodeString(String s) {
        StringBuilder res = new StringBuilder();
        Stack<Integer> multiple = new Stack<>();
        Stack<StringBuilder> stack = new Stack<>();
        int idx = 0;
        while(idx<s.length()){
            if(Character.isDigit(s.charAt(idx))){
                int count = 0;
                while(Character.isDigit(s.charAt(idx))){
                    count = count*10 + s.charAt(idx) - '0';
                    idx++;
                }
                multiple.push(count);
            } else if(s.charAt(idx) == '['){
                if(res.length()>0){
                    stack.push(res);
                    res = new StringBuilder();
                }
                idx++;
            }else if(s.charAt(idx) == ']'){
                int mult = multiple.pop();
                String cur = res.toString();
                while(mult>1){
                    res.append(cur);
                    mult--;
                }
                if(!stack.isEmpty()){
                    StringBuilder cursb = stack.pop();
                    cursb.append(res.toString());
                    res = cursb;
                }
                idx++;
                
            }else{
                res.append(s.charAt(idx));
                idx++;
            }
        }
        return res.toString();
    }
}

三刷
同上

class Solution {
    public String decodeString(String s) {
        StringBuilder res = new StringBuilder();
        Stack<Integer> multiple = new Stack<>();
        Stack<StringBuilder> base = new Stack<>();
        int idx = 0;
        while(idx<s.length()){
            char ch = s.charAt(idx);
            if(Character.isDigit(ch)){
                int count = 0;
                while(Character.isDigit(s.charAt(idx))){
                    count = 10*count + s.charAt(idx)-'0';
                    idx++;
                }
                multiple.push(count);
            }else if(ch == '['){
                if(res.length()!=0){
                    base.push(res);
                    res = new StringBuilder();
                }
                idx++;
            }else if(ch == ']'){
                int mul = multiple.pop();
                String cur = res.toString();
                while(mul>1){
                    res.append(cur);
                    mul--;
                }
                if(!base.isEmpty()) res = base.pop().append(res.toString());
                idx++;
            }else{
                res.append(ch);
                idx++;
            }
        }
        return res.toString();
    }
}