題目如下：

解題思路：我用的是遞歸的方法，每次找出與第一個‘)‘匹配的‘(‘計算atom的數量後去除括號，只到分子式中沒有括號為止。例如 "K4(ON(SO3)2)2" -> "K4(ONS2O6)2" -> "K4O2N2S4O12"。接下來再對分子式進行分割，得出每個atom的數量後排序即可。原理很簡單，代碼寫得很亂，僅供參考。

代碼如下：

class Solution(object):
    def recursive(self,formula):
        left = right = None
        for i,v in enumerate(formula):
            
 
if v == ‘(‘:
                left = i
            elif v == ‘)‘:
                right = i
                break
        if left == None and right == None:
            return formula
        lf = formula[:left]
        parse = formula[left+1:right]
        times = ‘‘
        for i in range(right+1,len(formula)):
            
 
if formula[i].isdigit():
                times += formula[i]
            else:
                if i != len(formula) - 1:
                    i -= 1
                break

        if times != ‘‘:
            times = int(times)

        rf = formula[i+1:]

        if times == ‘‘:
            ts = parse
        
 
else:
            parseList = []
            val = ‘‘
            val_num = ‘‘
            parse += ‘#‘
            for i in parse:
                #print parseList
                if i.islower():
                    val += i
                    #parseList.append(val)
                elif i.isupper():
                    if val != ‘‘:
                        parseList.append(val)
                    if val_num != ‘‘:
                        parseList.append(str(int(val_num) * int(times)))
                        val_num = ‘‘
                    elif val_num == ‘‘ and val != ‘‘:
                        parseList.append(str(times))
                    val = i
                elif i.isdigit():
                    if val != ‘‘:
                        parseList.append(val)
                        val = ‘‘
                    val_num += i
                elif i == ‘#‘:
                    if val != ‘‘:
                        parseList.append(val)
                    if val_num != ‘‘:
                        parseList.append(str(int(val_num) * int(times)))
                    elif val_num == ‘‘ and val != ‘‘:
                        parseList.append(str(times))
            ts = ‘‘.join(parseList)
        return self.recursive(lf + ts + rf)

    def countOfAtoms(self, formula):
        """
        :type formula: str
        :rtype: str
        """
        f =  self.recursive(formula)
        i = 1

        #print f

        #transform MgO2H2 -> Mg1O2H2
        while i < len(f):
            if f[i].isupper() and f[i-1].isdigit() == False:
                f = f[:i] + ‘1‘ + f[i:]
                i = 1
            i += 1
        if f[-1].isdigit() == False:
            f += ‘1‘

        dic = {}

        key = ‘‘
        val = ‘‘

        # H11He49N1O35B7N46Li20
        for i in f:
            if i.isdigit():
                val += i
            else:
                if val == ‘‘:
                    key += i
                else:
                    if key not in dic:
                        dic[key] = int(val)
                    else:
                        dic[key] += int(val)
                    key = i
                    val = ‘‘

        if key not in dic:
            dic[key] = int(val)
        else:
            dic[key] += int(val)


        keys = dic.keys()
        keys.sort()
        res = ‘‘
        #print dic
        for i in keys:
            res += i
            if dic[i] > 1:
                res += str(dic[i])
        return res
        

【leetcode】726. Number of Atoms