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

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

68ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         guard !A.isEmpty else {
 4             return []
 5         }
 6         var array = A
 7         var left = 0
 8         var right = A.count - 1           
 9         while(left < right && right > left) {
10             while(array[left] & 1 == 0 && left < right) {
11                 left += 1
12             }
13             while(array[right] & 1 == 1 && right > left) {
14                 right -= 1
15             }
16             if left != right {
17                 let temp = array[left]
18                 array[left] = array[right]
19                 array[right] = temp
20                 left += 1
21                 right -= 1
22             }
23         }  
24         return array
25     }
26 }

Runtime: 72 ms

Memory Usage: 19.4 MB

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         var A = A
 4         var left = 0
 5         var right = 0
 6         let count = A.count
 7         while right < count {
 8             // 偶数
 9             if A[right] & 1 == 0 { 
10                 if left != right {
11                     A[left] ^= A[right]
12                     A[right] = A[left] ^ A[right]
13                     A[left] ^= A[right]
14                 }
15                 left += 1
16             }
17             right += 1
18         }
19         return A
20     }
21 }

72ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         var even: [Int] = [], odd: [Int] = []
 4         A.forEach {
 5             if $0 % 2 == 0 {
 6                 even.append($0)
 7             } else {
 8                 odd.append($0)
 9             }
10         }
11         
12         return even + odd
13     }
14 }

76ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         var res = A
 4         var i = 0, pos = 0
 5     
 6         while i < A.count {
 7             if res[i] & 1 == 0 {
 8                 let val = res.remove(at: i)
 9                 res.insert(val, at: pos)
10                 pos += 1
11             }
12             i += 1
13         }
14     
15         return res
16     }
17 }

80ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         var parity: [Int] = []
 4         
 5         for num in A {
 6             if num & 0b1 == 0b1 {
 7                 parity.insert(num, at: parity.count)
 8             } else {
 9                 parity.insert(num, at: 0)
10             }
11         }
12         
13         return parity
14     }
15 }

84ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3     var arr = A
 4     var i = 0
 5     var j = A.count - 1
 6     while (i < j) {
 7         if (arr[i] % 2 > arr[j] % 2) {
 8             var temp = arr[i]
 9             arr[i] = arr[j]
10             arr[j] = temp
11         }
12         
13         if (arr[i] % 2 == 0) { i += 1 }
14         if (arr[j] % 2 == 1) { j -= 1 }
15     }
16     return arr
17     }
18 }

88ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         let odd = A.filter { (num) -> Bool in
 4             return (num % 2 == 1)
 5         }
 6         let even = A.filter { (num) -> Bool in
 7             return (num % 2 == 0)
 8         }
 9         return even + odd
10     }
11 }

100ms

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

104ms

1 class Solution {
2     func sortArrayByParity(_ A: [Int]) -> [Int] {
3         var B = A
4         _ = B.partition(by: { $0 % 2 != 0 })
5         return B
6     }
7 }

116ms

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

132ms

1 class Solution {
2     func sortArrayByParity(_ A: [Int]) -> [Int] {
3         return A.sorted(by: { (lhs, rhs) -> Bool in
4             return lhs % 2 == 0
5         })
6     }
7 }

184ms

 1 class Solution {
 2     func sortArrayByParity(_ A: [Int]) -> [Int] {
 3         return A.sorted { lhs, rhs in
 4             switch (lhs % 2, rhs % 2) {
 5             case (0, 0):
 6                 return true
 7             case (0, 1):
 8                 return true
 9             case (1, 0):
10                 return false
11             case (1, 1):
12                 return true
13             default:
14                 fatalError()    
15             }
16         }
17     }
18 }

248ms

1 class Solution {
2     func sortArrayByParity(_ A: [Int]) -> [Int] {
3         return A.filter{ $0 % 2 == 0 } + A.filter { $0 % 2 != 0 }
4     }
5 }

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

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