题目描述：

LeetCode 592. Fraction Addition and Subtraction

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change it to the format of fraction that has denominator 1. So in this case, 2 should be converted to 2/1.

Example 1:

Input:"-1/2+1/2"
Output: "0/1"

Example 2:

Input:"-1/2+1/2+1/3"
Output: "1/3"

Example 3:

Input:"1/3-1/2"
Output: "-1/6"

Example 4:

Input:"5/3+1/3"
Output: "2/1"

Note:

1. The input string only contains '0' to '9', '/', '+' and '-'. So does the output.
2. Each fraction (input and output) has format ±numerator/denominator. If the first input fraction or the output is positive, then '+' will be omitted.
3. The input only contains valid irreducible fractions, where the numerator and denominator of each fraction will always be in the range [1,10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
4. The number of given fractions will be in the range [1,10].
5. The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

题目大意：

求分数加减法算式的值，结果化为最简分数。

注意：

1. 输入字符串只包含0~9，+-/。输出亦然
2. 每个分数之前包含±号，若第一个分数为正数，省去+
3. 输入只包含有效的最简分数，分子分母范围[1, 10]。若分母为1，表示该分数实际上是一个整数。
4. 分数的个数范围[1, 10]
5. 分子分母的结果在32位带符号整数范围之内

解题思路：

求分母的最小公倍数，记为LO

将分子乘以 LO / 分母

对分子求和，记为HI

求LO和HI的最大公约数的绝对值，记为GCD

结果的分子分母分别为HI / GCD, LO / GCD

Python代码：

class Solution(object):
    def fractionAddition(self, expression):
        """
        :type expression: str
        :rtype: str
        """
        def gcd(a, b):
            if a < b: a, b = b, a
            while b:
                a, b = b, a % b
            return a

        def lcm(a, b):
            return a * b / gcd(a, b)

        part = ''
        fractions = []
        for c in expression:
            if c in '+-':
                if part: fractions.append(part)
                part = ''
            part += c
        if part: fractions.append(part)

        hi = [int(e.split('/')[0]) for e in fractions]
        lo = [int(e.split('/')[1]) for e in fractions]
        
        LO = reduce(lcm, lo)
        HI = sum(h * LO / l for h, l in zip(hi, lo))
        GCD = abs(gcd(LO, HI))

        return '%s/%s' % (HI / GCD, LO / GCD)

 

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