[Leetcode] 726. Number of Atoms 解题报告

题目：

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：

采用迭代+栈应该也可以实现，不过这里我们采用递归的方法实现，思路更简单。每当我们遇到'('的时候，就递归调用本函数解析“()”里面字符串，并将结果添加到当前的map中。而遇到')'则说明需要结束当前的递归调用，所以需要将当前的mp内的每个element的个数乘以‘)’后面所代表的出现次数，并且返回map。否则就解析当前出现的元素以及次数，并更新map。

需要注意的是，我们必须在递归函数中采用传引用的方式传入formula以及当前的索引，否则由于大量的拷贝操作会引起LTE。

代码：

class Solution {
public:
string countOfAtoms(string formula) {
int index = 0;
map<string, int> mp = parser(formula, index);
string res;
for (auto entry : mp) { // construct the result
res += entry.first + (entry.second == 1 ? "" : to_string(entry.second));
}
return res;
}
private:
map<string, int> parser(string& str, int& pos) {
map<string, int> res;
while (pos < str.size()) {
if (str[pos] == '(') { // parse recursively
pos++;
for (auto entry : parser(str, pos)) {
res[entry.first] += entry.second;
}
} else if (str[pos] == ')') { // finish the recursion
int s = ++pos;
while (pos < str.size() && isdigit(str[pos])) {
pos++;
}
int multiple = stoi(str.substr(s, pos - s));
for (auto entry : res) {
res[entry.first] *= multiple;
}
return res;
} else if (isupper(str[pos])) { // parse the elements in current layer
int s = pos++;
while (pos < str.size() && islower(str[pos])) {
++pos;
}
string elem = str.substr(s, pos - s);
s = pos;
while (pos < str.size() && isdigit(str[pos])) {
++pos;
}
string num = str.substr(s, pos - s);
res[elem] += num.empty() ? 1 : stoi(num);
}
}
return res;
}
};