https://leetcode.cn/problems/decode-string/submissions/

class Solution {
public:
    string getDigits(string &s, size_t &ptr) {
        string ret;
        while (isdigit(s[ptr])) {
            ret.push_back(s[ptr++]);
        }
        return ret;
    }

    string getString(vector <string> &v) {
        string ret;
        for (const auto &s: v) {
            ret += s;
        }
        return ret;
    }

    string decodeString(string s) {
        vector <string> stk;
        size_t ptr = 0;

        while (ptr < s.size()) {
            char cur = s[ptr];
            if (isdigit(cur)) {
                // 获取一个数字并进栈
                string digits = getDigits(s, ptr);
                stk.push_back(digits);
            } else if (isalpha(cur) || cur == '[') {
                // 获取一个字母并进栈
                stk.push_back(string(1, s[ptr++])); 
            } else {
                vector <string> sub;
                while (stk.back() != "[") {
                    sub.push_back(stk.back());
                    stk.pop_back();
                }
                reverse(sub.begin(), sub.end());
                // 左括号出栈
                stk.pop_back();
                // 此时栈顶为当前 sub 对应的字符串应该出现的次数
                int repTime = stoi(stk.back()); 
                stk.pop_back();
                string t, o = getString(sub);
                // 构造字符串
                while (repTime--) t += o; 
                // 将构造好的字符串入栈
                stk.push_back(t);
                //下一个字符
                ++ptr;
            }
        }

        return getString(stk);
    }
};


https://leetcode.cn/problems/compress-string-lcci/submissions/
class Solution {
public:
    string GetDigits(int count)
    {
        string tmp;
        while(count)
        {
            tmp+=count%10+'0';
            count/=10;
        }
        reverse(tmp.begin(), tmp.end());

        return tmp;
    }

    string compressString(string S) {
        string tmp; //存字符串
        int count=0;    //记录单个字母次数
        char ch=S[0];   //记录当前字母
        for(size_t i=0; i<=S.size(); ++i)   //i==S.size(), 便于最后一次存
        {
            if(ch != S[i])
            {
                tmp+=ch;
                tmp+=GetDigits(count);  //得到数字字符

                ch=S[i];   //更新
                count=1;
            }
            else
            {
                ++count;
            }
        }

        if(S.size()<=tmp.size())
        {
            return S;
        }

        return tmp;
    }
};


https://leetcode.cn/problems/multiply-strings/submissions/

class Solution {
public:

    string addStrings(string &num1, string &num2) 
    {
        int i = num1.size() - 1, j = num2.size() - 1, add = 0;
        string ans;
        while (i >= 0 || j >= 0 || add != 0) 
        {
            int x = i >= 0 ? num1[i] - '0' : 0;
            int y = j >= 0 ? num2[j] - '0' : 0;
            int result = x + y + add;
            ans+=result % 10+'0';
            add = result / 10;
            i--;
            j--;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }

    string multiply(string num1, string num2) {
        string ans = "0";
        if (num1 == "0" || num2 == "0") 
        {
            return ans;
        }
        
        int m = num1.size(), n = num2.size();
        for (int i = n - 1; i >= 0; i--) 
        {
            string curr;
            int add = 0;
            for (int j = n - 1; j > i; j--) //23*125    即十位2*125代表20*125.
            {
                curr+='0';
            }
            
            int y = num2[i] - '0';
            for (int j = m - 1; j >= 0; j--) 
            {
                int x = num1[j] - '0';
                int product = x * y + add;
                curr+=product % 10+'0';
                add = product / 10;
            }
            while (add != 0) 
            {
                curr+=add % 10+'0';
                add /= 10;
            }
            reverse(curr.begin(), curr.end());

            ans = addStrings(ans, curr);    //得出每一位的结果相加（例如：23*125即375与2400结果相加）
        }
        return ans;
    }
};