leetcode 726. Number of Atoms 化学式展开 + 统计化学式的元素的数量

Given a chemical formula (given as a string), return the count of each atom.

An atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name.

1 or more digits representing the count of that element may follow if the count is greater than 1. If the count is 1, no digits will follow. For example, H2O and H2O2 are possible, but H1O2 is impossible.

Two formulas concatenated together produce another formula. For example, H2O2He3Mg4 is also a formula.

A formula placed in parentheses, and a count (optionally added) is also a formula. For example, (H2O2) and (H2O2)3 are formulas.

Given a formula, output the count of all elements as a string in the following form: the first name (in sorted order), followed by its count (if that count is more than 1), followed by the second name (in sorted order), followed by its count (if that count is more than 1), and so on.

Example 1:
Input:
formula = “H2O”
Output: “H2O”
Explanation:
The count of elements are {‘H’: 2, ‘O’: 1}.
Example 2:
Input:
formula = “Mg(OH)2”
Output: “H2MgO2”
Explanation:
The count of elements are {‘H’: 2, ‘Mg’: 1, ‘O’: 2}.
Example 3:
Input:
formula = “K4(ON(SO3)2)2”
Output: “K4N2O14S4”
Explanation:
The count of elements are {‘K’: 4, ‘N’: 2, ‘O’: 14, ‘S’: 4}.
Note:

All atom names consist of lowercase letters, except for the first character which is uppercase.
The length of formula will be in the range [1, 1000].
formula will only consist of letters, digits, and round parentheses, and is a valid formula as defined in the problem.

就是直接解析字符串，然后获取元素的数量

其实我觉得直接按照括号匹配做展开也行，然后遍历统计元素也可以

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;



class Solution 
{
public:
    string countOfAtoms(string formula) 
    {
        int n = formula.size();
        map<string, int> stats;
        reverse(formula.begin(), formula.end());
        compute(formula, stats);
        string result;

        for (auto& kv : stats) 
        {
            result += kv.first;
            if (kv.second > 1)
                result += std::to_string(kv.second);
        }
        return result;
    }

     // K4(ON(SO3)2)2
    void compute(string& formula, std::map<string, int>& stats)
    {
        int n = formula.size();

        std::stack<int> stk; // factor
        stk.push(1);

        string elem; // chemical element, e.g., C, N, O, Na, Mg, Al, etc.
        int factor = 1; // number of this element, default to 1 if not otherwise specified

        for (int i = 0; i < n; ++i) 
        {
            auto& c = formula[i];
            if (isdigit(c)) 
            { // E.g., original 312 was reverse to 213.
                int val = 0, expo = 1;
                do 
                {
                    val += (c - '0') * expo;
                    ++i; 
                    expo *= 10;
                } while (isdigit(c = formula[i]));
                factor = val; // explicit number of this element
                i -= 1;
            }
            else if (c == ')')
            {
                stk.push(factor * stk.top());
                factor = 1; //            Al(OH)3
            }
            else if (c >= 'A' && c <= 'Z') 
            {
                elem += c;
                std::reverse(elem.begin(), elem.end());
                stats[elem] += factor * stk.top();
                factor = 1;
                elem.clear();
            }
            else if (c >= 'a' && c <= 'z')
            {
                elem += c;
            }
            else 
            {    // (c == '(') {  
                stk.pop();  // Ca(OH)2
            }
        }
    }
};