Given an encoded string, return it's decoded string.
The encoding rule is:k[encoded_string], where theencoded_stringinside the square brackets is being repeated exactlyktimes. Note thatkis 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 like3aor2[4].
Examples:
s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
用两个栈， 一个存数字， 另一个存 [ 之前产生的字符串，遇到 】 读栈顶，然后计算循环几次。
stack num： 3 ， 2
stack str： “” ，a，
】：a cc ； 】 “”acc acc acc
