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-rw-r--r--number_atoms.cpp75
-rw-r--r--number_atoms.py47
2 files changed, 122 insertions, 0 deletions
diff --git a/number_atoms.cpp b/number_atoms.cpp
new file mode 100644
index 0000000..2f05d3f
--- /dev/null
+++ b/number_atoms.cpp
@@ -0,0 +1,75 @@
+// 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
+}
diff --git a/number_atoms.py b/number_atoms.py
new file mode 100644
index 0000000..86fa2d7
--- /dev/null
+++ b/number_atoms.py
@@ -0,0 +1,47 @@
+# 726. Number of Atoms
+from collections import defaultdict
+
+"""
+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(object):
+ def countOfAtoms(self, formula):
+ """
+ :type formula: str
+ :rtype: str
+ """
+ dic, coeff, stack, elem, cnt, i = defaultdict(int), 1, [], "", 0, 0
+ for c in formula[::-1]:
+ if c.isdigit():
+ cnt += int(c) * (10**i)
+ i += 1
+ elif c == ")":
+ stack.append(cnt)
+ coeff *= cnt
+ i = cnt = 0
+ elif c == "(":
+ coeff //= stack.pop()
+ i = cnt = 0
+ elif c.isupper():
+ elem += c
+ dic[elem[::-1]] += (cnt or 1) * coeff
+ elem = ""
+ i = cnt = 0
+ elif c.islower():
+ elem += c
+ return "".join(k + str(v > 1 and v or "") for k, v in sorted(dic.items()))
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.countOfAtoms(formula="H2O"))
+ print(obj.countOfAtoms(formula="Mg(OH)2"))
+ print(obj.countOfAtoms(formula="K4(ON(SO3)2)2"))