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-rw-r--r--number_atoms.cpp75
1 files changed, 0 insertions, 75 deletions
diff --git a/number_atoms.cpp b/number_atoms.cpp
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--- a/number_atoms.cpp
+++ /dev/null
@@ -1,75 +0,0 @@
-// 726. Number of Atoms
-#include "leetcode.h"
-
-/*
- * given a string 'formula' representing a chemical formula, return the count of
- * each atom. the atomic element always stats with an uppercase character, then
- * zer or more lowercase letters, representing the name. one or more digits
- * representing that element's count may follow if the count is greater than 1.
- * if the count is 1, no digits will follow. return the count of all elements as
- * a string in the following form: the first name, followed by its count,
- * followed by the second name, followed by its count, and so on.
- */
-
-class Solution {
-public:
- string countOfAtoms(string formula) {
- string ans;
- const int n = formula.size();
- int i = 0;
- map<string, int> cnts = parse_formula(formula, i);
- for (pair<string, int> p : cnts) {
- ans += p.first;
- if (p.second > 1)
- ans += to_string(p.second);
- }
- return ans;
- }
-
-private:
- map<string, int> parse_formula(string &s, int &i) {
- map<string, int> cnts;
- const int n = s.size();
- while (i < n && s[i] != ')') {
- map<string, int> cnt = parse_unit(s, i);
- merge(cnts, cnt, 1);
- }
- return cnts;
- }
- map<string, int> parse_unit(string &s, int &i) {
- map<string, int> cnts;
- const int n = s.size();
- if (s[i] == '(') {
- map<string, int> cnt = parse_formula(s, ++i);
- int digits = parse_digits(s, ++i);
- merge(cnts, cnt, digits);
- } else {
- int i0 = i++;
- while (i < n && islower(s[i]))
- i++;
- string atom = s.substr(i0, i - i0);
- int digits = parse_digits(s, i);
- cnts[atom] += digits;
- }
- return cnts;
- }
- int parse_digits(string &s, int &i) {
- int i0 = i;
- while (i < s.size() && isdigit(s[i]))
- i++;
- int digits = i == i0 ? 1 : stoi(s.substr(i0, i - i0));
- return digits;
- }
- void merge(map<string, int> &cnts, map<string, int> &cnt, int times) {
- for (pair<string, int> p : cnt)
- cnts[p.first] += p.second * times;
- }
-};
-
-int main() {
- Solution obj;
- char f1[] = "H2O", f2[] = "Mg(OH)2", f3[] = "K4(ON(SO3)2)2";
- cout << obj.countOfAtoms(f1) << endl; // expect: H2O
- cout << obj.countOfAtoms(f2) << endl; // expect: H2MgO2
- cout << obj.countOfAtoms(f3) << endl; // expect: K4N2O14S4
-}