lintcode ：expression expand using C++

最新推荐文章于 2024-10-22 22:22:48 发布

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#lintcode #expressio #expand #string

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本文介绍了一种用于处理包含数字、字母及括号的表达式的字符串展开算法。该算法能够正确解析表达式中数字所指示的重复次数，并将括号内的字符串按指定次数重复输出。例如，输入3[abc]会输出abcabcabc。文章通过具体实例展示了算法的工作原理和实现过程。

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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

class Solution {
public:
    /**
    * @param s  an expression includes numbers, letters and brackets
    * @return a string
    */
    string expressionExpand(string& s) {
        // Write your code here
        string strOut = "";  // final string to be output
        string strTmp = "";  // the string between '[' and ']'
        stack<int> numOfStr; //store the number of the string to be expanded
        stack<char> strStore;//store the string input

        int strlen = s.length();
        int i = 0;
        while (i < strlen)
        {
            int count = 0;   //record the number
            if (s[i] == ']')
            {
                //get the string between []
                while (strStore.top() != '[')
                {
                    strTmp += strStore.top();
                    strStore.pop();

                //pop '['
                strStore.pop();

                int time = numOfStr.top();
                numOfStr.pop();
                string strTmp2(strTmp.rbegin(), strTmp.rend());
                int lenTmp = strTmp2.length();

                // expand the string and push it into the stack
                for (int j = 0; j < time; j++)
                {
                    for (int k = 0; k < lenTmp; k++)
                    {
                        strStore.push(strTmp2[k]);
                    }
                }
                strTmp = "";
                i++;

            }
            else if (isNum(s[i]) || s[i] == '[')
            {
                while (isNum(s[i]))
                {
                    //get int num from char num
                    int numtmp = s[i] - '0';
                    count = count * 10 + numtmp;
                    i++;
                }
                if (s[i] == '[')
                {
                    numOfStr.push(count);
                    strStore.push(s[i]);
                    i++;
                }

            }
            else
            {
                strStore.push(s[i]);
                i++;
            }
        }

        while (!strStore.empty())
        {
            strOut += strStore.top();
            strStore.pop();
        }
        //reverse the string because it is reversed as popped from stack 
        string strOut2(strOut.rbegin(), strOut.rend());
        return strOut2;
    }
    //check it is number or not
    bool isNum(char cr)
    {
        if (cr >= '0'&&cr <= '9')
            return true;
        else
            return false;
    }

};