题目描述：
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
formulawill be in the range[1, 1000]. formulawill only consist of letters, digits, and round parentheses, and is a valid formula as defined in the problem.
题目大意：
将分子式展开，返回各原子的个数
解题思路：
递归 + 字符串处理
Python代码：
class Solution(object):
def findGroup(self, formula):
cnt = 1
for x in range(1, len(formula)):
if formula[x] == '(': cnt += 1
elif formula[x] == ')': cnt -= 1
if cnt == 0: return formula[:x+1], formula[x+1:]
def findAtom(self, formula):
upper = 0
for x in range(len(formula)):
if formula[x].isupper(): upper += 1
if upper > 1 or not formula[x].isalpha():
return formula[:x], formula[x:]
return formula, ''
def findNum(self, formula):
for x in range(len(formula)):
if not formula[x].isdigit():
return int(formula[:x]), formula[x:]
return int(formula), ''
def count(self, formula):
cnt = collections.Counter()
while formula:
if formula[0] == '(':
group, formula = self.findGroup(formula)
groupCnt = self.count(group[1:-1])
num = 1
if formula and formula[0].isdigit():
num, formula = self.findNum(formula)
for k, v in groupCnt.iteritems():
cnt[k] += num * v
else:
atom, formula = self.findAtom(formula)
num = 1
if formula and formula[0].isdigit():
num, formula = self.findNum(formula)
cnt[atom] += num
return cnt
def countOfAtoms(self, formula):
"""
:type formula: str
:rtype: str
"""
cnt = self.count(formula)
return ''.join(k + str(cnt[k] > 1 and cnt[k] or '') for k in sorted(cnt))
本文链接：http://bookshadow.com/weblog/2017/11/13/leetcode-number-of-atoms/
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