給定一個(gè)表示分?jǐn)?shù)加減運(yùn)算表達(dá)式的字符串，你需要返回一個(gè)字符串形式的計(jì)算結(jié)果。 這個(gè)結(jié)果應(yīng)該是不可約分的分?jǐn)?shù)，即最簡(jiǎn)分?jǐn)?shù)。 如果最終結(jié)果是一個(gè)整數(shù)，例如 2，你需要將它轉(zhuǎn)換成分?jǐn)?shù)形式，其分母為 1。所以在上述例子中, 2 應(yīng)該被轉(zhuǎn)換為 2/1。

解法：分子 = a分子b分母 + a分母b分子 ，分母 = a分母 * b分母
然后約分，約分需要用到歐幾里得算法求最大公約數(shù)

func gcd(_ a:Int ,_ b:Int) -> Int {
    var inA = a,inB = b
    if a > b {
        inA = b
        inB = a
    }
    while inA != 0 {
        let temp = inA
        inA = inB % inA
        inB = temp
    }
    return abs(inB)
}
func fractionAddition(_ expression: String) -> String {
    
    var n: Int = 0,d: Int = 1
    var startIndex = expression.startIndex
    while startIndex < expression.endIndex {
        var leftIndex = startIndex
        repeat {
            leftIndex = expression.index(after: leftIndex)
        }while (expression[leftIndex] != "/");
        let range = expression[startIndex..<leftIndex]
        let fenzi:Int = Int(String(range))!
        var rightIndex = expression.index(after: leftIndex)
        while (rightIndex < expression.endIndex && expression[rightIndex] != "+" && expression[rightIndex] != "-") {
            rightIndex = expression.index(after: rightIndex)
        }
        let range2 = expression[expression.index(after: leftIndex)..<rightIndex]
        let fenmu:Int = Int(String(range2))!
        n = n*fenmu + fenzi*d
        d = d * fenmu
        let gcd = self.gcd(n, d)
        n = n / gcd
        d = d / gcd
        startIndex = rightIndex
    }
    return "\(n)/\(d)"
}