[Swift]LeetCode7. 反转整数 | Reverse Integer

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Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

---

给定一个 32 位有符号整数，将整数中的数字进行反转。

示例 1:

输入: 123
输出: 321

 示例 2:

输入: -123
输出: -321

示例 3:

输入: 120
输出: 21

注意:

假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−231,  231 − 1]。根据这个假设，如果反转后的整数溢出，则返回 0。

---

 12ms

class Solution {
    func reverse(_ x: Int) -> Int {
        //x为常量，赋值为变量
        var num:Int = x
        var rev:Int = 0
         while(num != 0)
        {
            var pop:Int = num % 10
            num /= 10
            if((rev > Int32.max/10)||(rev == Int32.max/10 && pop > 7)){return 0}
            if((rev < Int32.min/10)||(rev == Int32.min/10 && pop < -8)){return 0}
            rev = rev * 10 + pop
        }
        return rev
    }
}

---

12ms

class Solution {
    func reverse(_ x: Int) -> Int {
       
        guard x != 0  && abs(x) < INT32_MAX else{
            return 0
        }
        var result = 0
        var num = abs(x)
        while num > 0{
            let cursor = num % 10
            if cursor == 0 && result == 0{
                num /= 10
                continue
            }else{
                result = result * 10 + cursor
                num /= 10
            }
        }
        if result >= INT32_MAX{
            return 0
        }
        return (x > 0) ? result : -result
    }
}

---

16ms

class Solution {
    func reverse(_ x: Int) -> Int {
        var str = String(x)
        var sign = ""
        if x < 0 {
          sign = String(str.remove(at: str.startIndex)) 
        }
        
        let result = Int("(sign)(String(str.reversed()))") ?? 0
        
        let lowerLimit = Int(pow(Double(-2), 31))
        let higherLimit = Int(pow(Double(2), 31)-1)
        
        return result < lowerLimit || result > higherLimit ? 0 : result
    }
}

---

20ms

class Solution {
    func reverse(_ x: Int) -> Int {
      let str = Array(String(x))
        var rev : [Character] = []
        for i in 0..<str.count {
            let index = str.count-i-1
            if (str[index] == "-") { continue }
            rev.append(str[index])
        }
                
        // if we can convert the reverse string to a 32bit Int ... else return 0
        if let result = Int32(String(rev)) { 
        // signum == 1 or -1 if x is negative
            return Int(result) * x.signum() 
        } else {
            return 0
        }
}
}

---

24ms

class Solution {
    func reverse(_ x: Int) -> Int {
        let tmp = x < 0 ? -1 : 1
        let returnVal = Int(String(abs(x).description.reversed())) ?? 0
        if returnVal > Int(Int32.max) || returnVal < Int(Int32.min) {
            return 0
        }
        return returnVal * tmp
    }
}

---

28ms

class Solution {
    func reverse(_ x: Int) -> Int {
        var temp: Int = x
        var answer: UInt = 0
        var isNegative: Bool = false

        // Flip negatiave to positive
        if x < 0 {
            isNegative = true
            if x == Int.min {
                temp = (temp + 1) * -1
                temp += 1
            } else {
                temp *= -1
            }
        }
        
        // Reverse the numbers
        while temp > 0 {
            // Get last digit
            answer = (answer * 10) + (UInt(temp) % 10)
            
            // Remove last digit
            temp = temp / 10
        }
        
        print("answer: (answer)")
        print("max: (UInt(Int.max))")
        
        if answer > UInt(Int32.max) {
            return 0
        }
        
        if (isNegative) {
            return Int(answer) * -1
        } else {
            return Int(answer)
        }
    }
}