lintcode--Expression Expand

题目：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.

样例

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

解答：解答本题有两点：（1）用栈做DFS，注意栈不为空，并且栈是先进后出的，注意顺序；（2）numbers可能是多位数。

class Solution {public:    /**     * @reverse the word, include words and numbers     */    void reverseWord(string &s) {        for (int i = 0, j = s.size() - 1; i < j; i++,j--) {            swap(s[i], s[j]);        }    }    /**     * @param s  an expression includes numbers, letters and brackets     * @return a string     */    string expressionExpand(string& s) {        // Write your code here        stack<char> res;        string val = "";        int len = s.length();        if (len == 0) {            return val;        }        for (int i = 0; i < len; i++) {            if (s[i] == ']') {                string tmp = "";                while (!res.empty()) {                    char ch = res.top();                    res.pop();                    if (ch == '[') {                        break;                    }                    tmp += ch;                }                string num = "";                while (!res.empty()) {                    char tmpnum = res.top();                    res.pop();                    if(isdigit(tmpnum)){                        num += tmpnum;                    } else {                        res.push(tmpnum);                        break;                    }                }                reverseWord(num);                int count = atoi(num.c_str());                string mid = "";                reverseWord(tmp);                for (int j = 0; j <count; j++) {                    mid += tmp;                }                for (int j = 0; j < mid.length(); j++) {                    res.push(mid[j]);                }            }            else {                res.push(s[i]);            }        }                while (!res.empty()) {            val += res.top();            res.pop();        }        reverseWord(val);        return val;    }};