求解一個給定的方程，將x以字符串"x=#value"的形式返回。該方程僅包含'+'，' - '操作，變量 x 和其對應(yīng)系數(shù)。
如果方程沒有解，請返回“No solution”。
如果方程有無限解，則返回“Infinite solutions”。
如果方程中只有一個解，要保證返回值 x 是一個整數(shù)。

解析字符串，按照普通的解方程手法進(jìn)行

let index:String.Index = equation.index(of: "=")!
    let leftStr = equation[..<index]
    let rightStr = equation[equation.index(after: index)...]
    var leftNum: Int = 0
    var leftX: Int = 0
    var tempStr: String = ""
    for aChar in leftStr {
        if aChar == "x" {
            if tempStr.isEmpty {
                leftX += 1
            }else {
                if tempStr.last == "+" || tempStr.last == "-" {
                    tempStr += "1"
                }
                leftX += Int(tempStr)!
                tempStr = ""
            }
        }else if aChar == "+" || aChar == "-" {
            if !tempStr.isEmpty {
                leftNum += Int(tempStr)!
            }
            tempStr = String(aChar)
        }else {
            tempStr += String(aChar)
        }
    }
    if !tempStr.isEmpty {
        leftNum += Int(tempStr)!
        tempStr = ""
    }
    var rightNum: Int = 0
    var rightX: Int = 0
    for aChar in rightStr {
        if aChar == "x" {
            if tempStr.isEmpty {
                rightX += 1
            }else {
                if tempStr.last == "+" || tempStr.last == "-" {
                    tempStr += "1"
                }
                rightX += Int(tempStr)!
                tempStr = ""
            }
        }else if aChar == "+" || aChar == "-" {
            if !tempStr.isEmpty {
                rightNum += Int(tempStr)!
            }
            tempStr = String(aChar)
        }else {
            tempStr += String(aChar)
        }
    }
    if !tempStr.isEmpty {
        rightNum += Int(tempStr)!
    }
    leftX = leftX - rightX
    rightNum = rightNum - leftNum
    var rstStr = ""
    if leftX == 0 {
        if rightNum == 0 {
            rstStr = "Infinite solutions"
        }else {
            rstStr = "No solution"
        }
    }else {
        let rstNum = Double(rightNum) / Double(leftX)
        var rstInt = Int(rstNum)
        if rstNum < Double(rstInt) {
            rstInt -= 1
        }
        rstStr = "x=\(rstInt)"
    }
    return rstStr