[Swift]LeetCode905. 按奇偶排序数组 | Sort Array By Parity

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Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition. 

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted. 

Note:

1. 1 <= A.length <= 5000
2. 0 <= A[i] <= 5000

---

给定一个非负整数数组 A，返回一个由 A 的所有偶数元素组成的数组，后面跟 A 的所有奇数元素。

你可以返回满足此条件的任何数组作为答案。 

示例：

输入：[3,1,2,4]
输出：[2,4,3,1]
输出 [4,2,3,1]，[2,4,1,3] 和 [4,2,1,3] 也会被接受。 

提示：

1. 1 <= A.length <= 5000
2. 0 <= A[i] <= 5000

---

68ms

 

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        guard !A.isEmpty else {
            return []
        }
        var array = A
        var left = 0
        var right = A.count - 1           
        while(left < right && right > left) {
            while(array[left] & 1 == 0 && left < right) {
                left += 1
            }
            while(array[right] & 1 == 1 && right > left) {
                right -= 1
            }
            if left != right {
                let temp = array[left]
                array[left] = array[right]
                array[right] = temp
                left += 1
                right -= 1
            }
        }  
        return array
    }
}

---

Runtime: 72 ms

Memory Usage: 19.4 MB

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        var A = A
        var left = 0
        var right = 0
        let count = A.count
        while right < count {
            // 偶数
            if A[right] & 1 == 0 { 
                if left != right {
                    A[left] ^= A[right]
                    A[right] = A[left] ^ A[right]
                    A[left] ^= A[right]
                }
                left += 1
            }
            right += 1
        }
        return A
    }
}

---

72ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        var even: [Int] = [], odd: [Int] = []
        A.forEach {
            if $0 % 2 == 0 {
                even.append($0)
            } else {
                odd.append($0)
            }
        }
        
        return even + odd
    }
}

---

76ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        var res = A
        var i = 0, pos = 0
    
        while i < A.count {
            if res[i] & 1 == 0 {
                let val = res.remove(at: i)
                res.insert(val, at: pos)
                pos += 1
            }
            i += 1
        }
    
        return res
    }
}

---

80ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        var parity: [Int] = []
        
        for num in A {
            if num & 0b1 == 0b1 {
                parity.insert(num, at: parity.count)
            } else {
                parity.insert(num, at: 0)
            }
        }
        
        return parity
    }
}

---

84ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
    var arr = A
    var i = 0
    var j = A.count - 1
    while (i < j) {
        if (arr[i] % 2 > arr[j] % 2) {
            var temp = arr[i]
            arr[i] = arr[j]
            arr[j] = temp
        }
        
        if (arr[i] % 2 == 0) { i += 1 }
        if (arr[j] % 2 == 1) { j -= 1 }
    }
    return arr
    }
}

---

88ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        let odd = A.filter { (num) -> Bool in
            return (num % 2 == 1)
        }
        let even = A.filter { (num) -> Bool in
            return (num % 2 == 0)
        }
        return even + odd
    }
}

---

100ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        return A.sorted(by: { $0 % 2 < $1 % 2 })
    }
}

---

104ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        var B = A
        _ = B.partition(by: { $0 % 2 != 0 })
        return B
    }
}

---

116ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        return A.sorted(by: { return $0 % 2 == 0 && $1 % 2 == 1})
    }
}

---

132ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        return A.sorted(by: { (lhs, rhs) -> Bool in
            return lhs % 2 == 0
        })
    }
}

---

184ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        return A.sorted { lhs, rhs in
            switch (lhs % 2, rhs % 2) {
            case (0, 0):
                return true
            case (0, 1):
                return true
            case (1, 0):
                return false
            case (1, 1):
                return true
            default:
                fatalError()    
            }
        }
    }
}

---

248ms

class Solution {
    func sortArrayByParity(_ A: [Int]) -> [Int] {
        return A.filter{ $0 % 2 == 0 } + A.filter { $0 % 2 != 0 }
    }
}