[Swift]LeetCode922.按奇偶排序数组 II | Sort Array By Parity II

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Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

 Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

 Note:

1. 2 <= A.length <= 20000
2. A.length % 2 == 0
3. 0 <= A[i] <= 1000

---

给定一个非负整数数组 A， A 中一半整数是奇数，一半整数是偶数。

对数组进行排序，以便当 A[i] 为奇数时，i 也是奇数；当 A[i] 为偶数时， i 也是偶数。

你可以返回任何满足上述条件的数组作为答案。

 示例：

输入：[4,2,5,7]
输出：[4,5,2,7]
解释：[4,7,2,5]，[2,5,4,7]，[2,7,4,5] 也会被接受。

 提示：

1. 2 <= A.length <= 20000
2. A.length % 2 == 0
3. 0 <= A[i] <= 1000

---

288ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        guard A.count > 1 else {
            return A
        }
        
        var A = A
        var evenPointer = 0
        var oddPointer = 1
        while evenPointer < A.count {
            
            if A[evenPointer] % 2 == 0 {
                evenPointer += 2
            } else {
                A.swapAt(evenPointer, oddPointer)
                oddPointer += 2
            }
        }
        return A
    }
}

---

292ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        let n = A.count
        var A = A
        var i = 0
        var j = 1
        while i < n && j < n {
            while i < n && A[i] % 2 == 0 {
                i += 2
            }
            while (j < n && A[j] % 2 == 1) {
                j += 2
            }
            if i < n && j < n {
                (A[i], A[j]) = (A[j], A[i])
            }
        }
        return A;
    }
}

---

296ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        var outputEven: [Int] = []
        var outputOdd: [Int] = []
        
        for value in A {
            if value % 2 == 0 {
                outputEven.append(value)
            } else {
                outputOdd.append(value)
            }
        }

        var output:[Int] = []
        for index in 0 ..< outputEven.count {
            output.append(outputEven[index])
            output.append(outputOdd[index])
        }
        return output
    }
}

---

300ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        
        var even = [Int]()
        var odd = [Int]()
        
        for i in A {
            if i % 2 == 0 {
                even.append(i)
            } else {
                odd.append(i)
            }
        }
        
        var output = [Int]()
        
        for i in 0..<even.count {
            output.append(even[i])
            output.append(odd[i])
        }
        
        return output
    }
}

---

308ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        var copyA = A
        var oddArray: [Int] = []
        var evenArray: [Int] = []
        for (index, interger) in A.enumerated() {
            if index % 2 == 0, interger % 2 != 0 {
                oddArray.append(index)
            }
            if index % 2 != 0, interger % 2 == 0 {
                evenArray.append(index)
            }
        }
        for (index, _) in oddArray.enumerated() {
            copyA.swapAt(oddArray[index], evenArray[index])
        }
        return copyA
    }
}

---

320ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        var evenNums = A.filter{$0 % 2 != 0}
        var oddNums = A.filter{$0 % 2 == 0}
        var result = [Int]()
        
        for (index , _) in oddNums.enumerated() {
            result.append(oddNums[index])
            result.append(evenNums[index])
        }
        
        return result
    }
}

---

324ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        var N:Int = A.count
        var ans:[Int] = [Int](repeating: 0,count: N)
        
        var t:Int = 0
        for x in A
        {
            if x % 2 == 0
            {
                ans[t] = x
                t += 2
            }         
        }
        t = 1
        for x in A
        {
             if x % 2 == 1
            {
                ans[t] = x
                t += 2
            }
        }
        return ans
    }
}

---

328ms

class Solution {
    func sortArrayByParityII(_ a: [Int]) -> [Int] {
        var evenNumbers: [Int] = []
        var oldNumbers: [Int] = []
        for n in a {
            if n % 2 == 0 {
                evenNumbers.append(n)
            } else {
                oldNumbers.append(n)
            }
        }
        
        let zipped = zip(evenNumbers,oldNumbers).flatMap{ [$0.0, $0.1] }
        return zipped
    }
}

---

368ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        var res1 = [Int]()
        var res2 = [Int]()
        var res3 = [Int]()

        for num in A {
            if num % 2 == 0{//ou
                res1.append(num)
            }
            else{//ji
                res2.append(num)
            } 
        }
        for i in 0..<res1.count{
            res3.append(res1[i])
            res3.append(res2[i])
        }
        return res3
    }
}

---

416ms

class Solution {
    func sortArrayByParityII(_ A: [Int]) -> [Int] {
        let b = A.sorted { (i, _) in return i % 2 == 0 }
        let count = A.count
        var result:[Int] = Array(repeating: 0, count: count)
        for i in 0..<count/2 {
            result[2*i] = b[i]
            result[2*i+1] = b[count-1-i]
        }

        return result
    }
}