题目如下：

解题思路：本题考察的是分数的加减法。小学时候就学过，分数的加减法是先求两个分母的最小公倍数，然后分子分别乘以最小公倍数与自己分母的商，相加后约分即可。所以，本题只要按+,-两个符号分割输入字符串，就可以得到所有的分数列表，做加减操作即可。考虑到第一个分数是负数的情况，我在代码中加入了一个判断，如果表达式第一个字符是负数，给表达式加上'0/1'的前缀，既不影响结果，又简化了代码。

代码如下：

class Solution(object):    def fractionAddition(self, expression):        """        :type expression: str        :rtype: str        """        def gcd(a, b):            if a % b == 0:                return b            else:                return gcd(b, a % b)        def lcm(a,b):            return  a * b / gcd(a,b)        if expression[0] == '-':            expression = '0/1' + expression        expression += '#'        operators = ['+','-']        operator = ''        next = ''        res = ''        for i in expression:            if i in operators or i == '#':                if operator == '':                    operator = i                    res = next                    next = ''                else:                    fl = res.split('/')                    sl = next.split('/')                    denominator = lcm(int(fl[1]),int(sl[1]))                    if operator == '+':                        numerator = int(fl[0]) *  denominator/int(fl[1])  + int(sl[0]) *  denominator/int(sl[1])                    else:                        numerator = int(fl[0]) * denominator / int(fl[1]) - int(sl[0]) * denominator / int(sl[1])                    res = str(numerator) + '/' + str(denominator)                    next = ''                    operator = i            else:                next += i        rl = res.split('/')        g = gcd(int(rl[0]),int(rl[1]))        res = str(int(rl[0])/g) + '/' +  str(int(rl[1])/g)        return res