[Swift]LeetCode640. 求解方程 | Solve the Equation

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Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient.

If there is no solution for the equation, return "No solution".

If there are infinite solutions for the equation, return "Infinite solutions".

If there is exactly one solution for the equation, we ensure that the value of x is an integer.

Example 1:

Input: "x+5-3+x=6+x-2"
Output: "x=2" 

Example 2:

Input: "x=x"
Output: "Infinite solutions" 

Example 3:

Input: "2x=x"
Output: "x=0" 

Example 4:

Input: "2x+3x-6x=x+2"
Output: "x=-1" 

Example 5:

Input: "x=x+2"
Output: "No solution"

---

求解一个给定的方程，将x以字符串"x=#value"的形式返回。该方程仅包含'+'，' - '操作，变量 x 和其对应系数。

如果方程没有解，请返回“No solution”。

如果方程有无限解，则返回“Infinite solutions”。

如果方程中只有一个解，要保证返回值 x 是一个整数。

示例 1：

输入: "x+5-3+x=6+x-2"
输出: "x=2"

示例 2:

输入: "x=x"
输出: "Infinite solutions"

示例 3:

输入: "2x=x"
输出: "x=0"

示例 4:

输入: "2x+3x-6x=x+2"
输出: "x=-1"

示例 5:

输入: "x=x+2"
输出: "No solution"

---

Runtime: 4 ms

Memory Usage: 20 MB

class Solution {
    func solveEquation(_ equation: String) -> String {
        var arr:[Character] = Array(equation)
        var n:Int = equation.count
        var a:Int = 0
        var b:Int = 0
        var sign:Int = 1
        var j:Int = 0
        for i in 0..<n
        {
            if arr[i] == "+" || arr[i] == "-"
            {
                if i > j
                {
                    b += Int(equation.subString(j, i - j))! * sign
                }
                j = i
            }
            else if arr[i] == "x"
            {
                if i == j || arr[i - 1] == "+"
                {
                    a += sign
                }
                else if arr[i - 1] == "-"
                {
                    a -= sign
                }
                else
                {
                    a += Int(equation.subString(j, i - j))! * sign
                }
                j = i + 1
            }
            else if arr[i] == "="
            {
                if i > j
                {
                    b += Int(equation.subString(j, i - j))! * sign
                }
                sign = -1
                j = i + 1
            }
        }
        if j < n {b += Int(equation.subString(j))! * sign}
        if a == 0 && a == b {return "Infinite solutions"}
        if a == 0 && a != b {return "No solution"}
        return "x=" + String(-b / a)
    }
}

extension String {    
    // 截取字符串：指定索引和字符数
    // - begin: 开始截取处索引
    // - count: 截取的字符数量
    func subString(_ begin:Int,_ count:Int) -> String {
        let start = self.index(self.startIndex, offsetBy: max(0, begin))
        let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
        return String(self[start..<end]) 
    }
    
    // 截取字符串：从index到结束处
    // - Parameter index: 开始索引
    // - Returns: 子字符串
    func subString(_ index: Int) -> String {
        let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
        return String(self[theIndex..<endIndex])
    }
}

---

8ms

class Solution {
    func solveEquation(_ equation: String) -> String {
        let parts = equation.components(separatedBy: "=")
        let res1 = resolve(parts[0]), res2 = resolve(parts[1])
        let n = (res2.1 - res1.1)
        let x = (res1.0 - res2.0)
        if n == 0 && x == 0 { return "Infinite solutions" }
        if x == 0 { return "No solution" }
        return "x=\(n/x)"
    }
    
    private func resolve(_ e: String) -> (Int, Int) {
        var x = 0, n = 0
        var num = 0
        var isX = false
        var sign = 1
        var ca = Array(e)
        ca.append("+")
        for i in 0..<ca.count {
            let c = ca[i]
            if let digit = Int(String(c)) {
                num *= 10
                num += digit
            } else if c == "+" || c == "-" {
                if isX {
                    if i <= 1 || ca[i-2] != "0" {
                        num = num == 0 ? 1 : num
                    }
                    x += num * sign
                } else {
                    n += num * sign
                }
                sign = c == "+" ? 1 : -1
                num = 0
                isX = false
            } else if c == "x" {
                isX = true
            }
        }
        return (x, n)
    }
}