leetcode 726. Number of Atoms

726. Number
of Atoms

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.
思路和394. Decode
String一样，就是从最里层找起，找第一个')'，肯定是最里层的结束。

然后找到最里层的字符串，然后用map来存储每一个元素。最后再组装回去。

struct compare{
bool operator()(const string& a, const string& b) const
{
int i = 0;
while (i < a.size() && i < b.size() && a[i] == b[i])
i++;
return a[i] < b[i];
}
};

class Solution {
public:
string helper2(string s, int num) //SO3 2 返回 S2O6
{
map<string, int, compare> mp; //map自定义排序顺序
int i = 0;
while (i < s.size())
{
if (s[i] >= 'A' && s[i] <= 'Z') //开始
{
if (i + 1 == s.size() || (s[i + 1] >= 'A' && s[i + 1] <= 'Z'))
{
mp[s.substr(i, 1)] += num;
i++;
}
else
{
int j = i + 1;
if (s[i + 1] >= 'a' && s[i + 1] <= 'z') j ++;
string now = s.substr(i, j - i);
i = j;
while (j < s.size() && s[j] >= '0' && s[j] <= '9') j++;
int now_num = atoi(s.substr(i, j - i).c_str());
mp[now] += (now_num == 0) ? num : num * now_num; //now_num = 0 说明后面是没数字
i = j;
}
}
}
string ret = "";
for (auto it = mp.begin(); it != mp.end(); it++)
{
ret = ret + it->first;
if (it->second != 1)
ret = ret + to_string(it->second);
}
return ret;
}

string helper(string s, int i) //找到第一个')'然后往前找'('和往后找'数字',然后更改返回
{
int j = i - 1;
while (s[j] != '(') j--;
int k = i + 1;
while (s[k] >= '0' && s[k] <= '9') k++;

string ret = s.substr(0, j);
int num = atoi(s.substr(i + 1, k - i - 1).c_str());
ret = ret + helper2(s.substr(j + 1, i - j - 1), num);
ret = ret + s.substr(k);
return ret;
}

string countOfAtoms(string s)
{
int pos = s.find(')');
if (pos < 0 || pos >= s.size()) return helper2(s, 1); //最后还要处理一遍顺序
return countOfAtoms(helper(s, pos));
}
};