Leetcode題目[1]：
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

題目解釋：給定一個數組，將偶數的元素排在前面，奇數的元素排在後面。
解法一：
思路：將陣列元組逐個元素判斷奇偶性，然後偶性元素排在前面，奇數元素排在後面。

class Solution:
    def sortArrayByParity(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        odd_result = []
        even_result = []
        results = []
        for i in A:
            if i%2 == 0:
                even_result.append(i)
            if i%2 == 1:
                odd_result.append(i)
        for i in even_result:
            results.append(i)
        for i in odd_result:
            results.append(i)
        return results
 

解法二[2]：

class Solution(object):
    def sortArrayByParity(self, A):
        A.sort(key = lambda x: x % 2)
        return A

Complexity Analysis

    Time Complexity: O(NlogN), where N is the length of A.

    Space Complexity: O(N) for the sort, depending on the built-in implementation of sort. 
 

解法三、

class Solution(object):
    def sortArrayByParity(self, A):
        return ([x for x in A if x % 2 == 0] +
                [x for x in A if x % 2 == 1])

Complexity Analysis

    Time Complexity: O(N), where N is the length of A.

    Space Complexity: O(N) for the sort, depending on the built-in implementation of sort. 

解法三、

Approach 3: In-Place

Intuition

If we want to do the sort in-place, we can use quicksort, a standard textbook algorithm.

Algorithm

We'll maintain two pointers i and j. The loop invariant is everything below i has parity 0 (ie. A[k] % 2 == 0 when k < i), and everything above j has parity 1.

Then, there are 4 cases for (A[i] % 2, A[j] % 2):

    If it is (0, 1), then everything is correct: i++ and j--.

    If it is (1, 0), we swap them so they are correct, then continue.

    If it is (0, 0), only the i place is correct, so we i++ and continue.

    If it is (1, 1), only the j place is correct, so we j-- and continue.

Throughout all 4 cases, the loop invariant is maintained, and j-i is getting smaller. So eventually we will be done with the array sorted as desired.

class Solution(object):
    def sortArrayByParity(self, A):
        i, j = 0, len(A) - 1
        while i < j:
            if A[i] % 2 > A[j] % 2:
                A[i], A[j] = A[j], A[i]

            if A[i] % 2 == 0: i += 1
            if A[j] % 2 == 1: j -= 1

        return A

複雜度分析：

Complexity Analysis

    Time Complexity: O(N), where N is the length of A. Each step of the while loop makes j-i decrease by at least one. (Note that while quicksort is O(NlogN) normally, this is O(N) because we only need one pass to sort the elements.)

    Space Complexity: O(1) in additional space complexity. 

Ref:
1、https://leetcode.com/problems/sort-array-by-parity/description/
2、https://leetcode.com/problems/sort-array-by-parity/solution/#