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".

 这道题感觉挺难的，迭代的思想感觉不太容易想

1. Iterate through the entire string (while loop and idx++)
2. if the current char is number (isDigit()), convert string to int
3. if the current char is ‘[’(next is a char ), push res to strStack, clean res
4. if the current char is ‘]’ (build the res string), append res to previous string in strStack
5. if the current char is letter (we don’t know what’s next), add it to res

 

class Solution {    public String decodeString(String s) {        if(s == null || s.length() == 0){            return "";        }        Stack
     
       numStack = new Stack<>();        Stack
      
        strStack = new Stack<>();        int count = 0;        StringBuilder res = new StringBuilder();        for(int i =0; i< s.length(); i++){            char c = s.charAt(i);            //find number            if(c >= '0' && c <= '9'){                count = 10 * count + c - '0';            }            //find left square brackets            else if(c == '['){                numStack.push(count);                strStack.push(res.toString());                count = 0;                res = new StringBuilder();            }            //find right square brackets            else if( c == ']'){                StringBuilder temp = new StringBuilder(strStack.pop());                int repeat = numStack.pop();                for(int j = 0; j < repeat; j++){                    temp.append(res);                }                res = temp;            }            //find normal character            else{                res.append(c);            }        }        return res.toString();            }}