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".

这里需要分成4种情况考虑：

1、ch = 数字，向后遍历直到没有数字，把数字推入count堆栈

2、ch = [，string堆栈推入空字符串“”，开启新一篇

3、ch = ]， count出栈，stringbuilder.append() n 次

4、ch = 字母，result.push(result.pop() + ch)，这个设计很巧妙

代码如下：

public class Solution {
    public String decodeString(String s) {
        Stack<Integer> count = new Stack<Integer>();
        Stack<String> result = new Stack<String>();
        result.push("");
        int i = 0, len = s.length();
        while (i < len) {
            char ch = s.charAt(i); 
            if (ch >= '0' && ch <= '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 (ch == '[') {
                result.push("");
            } else if (ch == ']') {
                int n = count.pop();
                StringBuilder builder = new StringBuilder();
                String temp = result.pop();
                while (n > 0) {
                    builder.append(temp);
                    n --;
                }
                result.push(result.pop() + builder.toString());
            } else {
                result.push(result.pop() + ch);
            }
            i ++;
        }
        return result.pop();
    }
}