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".

Tips:

先建立count和result栈，注意result要先入栈"";注意每次result要push的情况下可能要 result.push(result.pop() + 当前要入栈的元素)

流程：
1.遇到数字换算一下存入count栈；
2.遇到[，result入栈“”；
3.遇到字母，result入栈字母；result.push(result.pop() + s.charAt(i))；
4.遇到]，将result.pop()入栈count.pop()次；
5.return result.pop()。

Code:

public class Solution {
    public String decodeString(String s) {
        Stack<Integer> count = new Stack<>();
        Stack<String> result = new Stack<>();
        result.push("");
        int i = 0;
        while (i < s.length()) {
           if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                int start = i;
                while (s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') {
                    i++;
                }
                count.push(Integer.parseInt(s.substring(start, i + 1)));
            } else if (s.charAt(i) == '[') {
                result.push("");
            } else if (s.charAt(i) == ']') {
                String temp = result.pop();
                StringBuilder sb = new StringBuilder();
                int index = count.pop();
                for (int j = 0; j < index; j++) {
                    sb.append(temp);
                }
                result.push(result.pop() + sb.toString());
            } else {
                result.push(result.pop() + s.charAt(i));
            }
            i++;
        }
        return result.pop();
    }
}