Given an expression s includes numbers, letters and brackets. Number represents the number of repetitions inside the brackets(can be a string or another expression)．Please expand expression to be a string.

s = abc3[a] return abcaaa

s = 3[abc] return abcabcabc

s = 4[ac]dy, return acacacacdy

s = 3[2[ad]3[pf]]xyz, return adadpfpfpfadadpfpfpfadadpfpfpfxyz

解题思路：把数字加入stack,遇到“["，把[]里的字母加入stack, 遇到"]"以后pop out字母，然后根据数字来决定最终字母的重复次数，pop out 数字，把整理好的字母，重新加入stack, 最后把所有stack里的字母拿出。

class elemType {
public:
    int num;
    string str = "";
    elemType(int num) {
        this->num = num;
    }
    elemType(string str) {
        this->str = str;
    }
};

class Solution {
public:

    string expressionExpand(string& s) {
        // Write your code here
        //if (s.size() == 0) {
        //    return NULL;
        //} 
        int number = 0;
        for (char c : s) {
            if (isdigit(c)) {
                number = number * 10 + (c - '0');
            } else if (c == '[') {
                elem.push(elemType(number));
                number = 0;
            } else if (c == ']') {
                string popString = stackPop();
                int count = elem.top().num;
                //pop out number;
                elem.pop();
                for (int i = 0; i < count; i++) {
                    elem.push(elemType(popString));
                }

            } else {
                elem.push(elemType(string(1,c)));
            }
        }
        string result = stackPop();
        return result;

    }
    string stackPop() {
        string temp = "";
        while (!elem.empty() && elem.top().str != "") {
            temp = elem.top().str + temp;
            elem.pop();
        }
        return temp;
    }
private:
    stack <elemType> elem;
};