394. Decode String

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 countStack = new Stack<>();
        Stack resStack = new Stack<>();
        int idx = 0;
        while(idx0){
                    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 multiple = new Stack<>();
        Stack stack = new Stack<>();
        int idx = 0;
        while(idx0){
                    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 multiple = new Stack<>();
        Stack base = new Stack<>();
        int idx = 0;
        while(idx1){
                    res.append(cur);
                    mul--;
                }
                if(!base.isEmpty()) res = base.pop().append(res.toString());
                idx++;
            }else{
                res.append(ch);
                idx++;
            }
        }
        return res.toString();
    }
}