[Swift]LeetCode1017. 负二进制转换 | Convert to Base -2

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Given a number N, return a string consisting of "0"s and "1"s that represents its value in base -2 (negative two).

The returned string must have no leading zeroes, unless the string is "0". 

Example 1:

Input: 2
Output: "110"
Explantion: (-2) ^ 2 + (-2) ^ 1 = 2

Example 2:

Input: 3
Output: "111"
Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3

Example 3:

Input: 4
Output: "100"
Explantion: (-2) ^ 2 = 4 

Note:

1. 0 <= N <= 10^9

---

给出数字 N，返回由若干 "0" 和 "1"组成的字符串，该字符串为 N 的负二进制（base -2）表示。

除非字符串就是 "0"，否则返回的字符串中不能含有前导零。 

示例 1：

输入：2
输出："110"
解释：(-2) ^ 2 + (-2) ^ 1 = 2

示例 2：

输入：3
输出："111"
解释：(-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3

示例 3：

输入：4
输出："100"
解释：(-2) ^ 2 = 4 

提示：

1. 0 <= N <= 10^9

---

8ms

class Solution {
    func baseNeg2(_ N: Int) -> String {
        var num = N
        var str = ""
        var base = -2
        while num < 0 || num >= 2 {
            var q = Int(num / base)
            if num < 0, num % base != 0 { q += 1 }
            str = "((num - q * base))(str)"
            num = q
        }
        if num != 0 {
            str = "(num)(str)"
        }
        if str.isEmpty {
            str = "0"
        }
        return str
    }
}

---

Runtime: 16 ms

Memory Usage: 18.9 MB

class Solution {
    func baseNeg2(_ N: Int) -> String {
        var N = N
        var ans:String = String()
        while(N != 0)
        {
            var r:Int = abs(N % 2)
            if r != 0 { N -= 1}
            var q:Int = N / -2
            ans.append((r + 48).ASCII)
            N = q            
        }
        ans = String(ans.reversed())
        if ans.isEmpty
        {
            ans.append("0")
        }
        return ans        
    }
}

//Int扩展
extension Int
{
    //Int转Character,ASCII值(定义大写为字符值)
    var ASCII:Character 
    {
        get {return Character(UnicodeScalar(self)!)}
    }
}