LeetCode 905. Sort Array By Parity

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题目

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:

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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

思路

  • Approach 1: Sort
    Use a custom comparator when sorting, to sort by parity.
    Time Complexity: O(NlogN), Space Complexity: O(N)
  • Approach 2: Two Pass
    Write all the even elements first, then write all the odd elements.
    Time Complexity: O(N), Space Complexity: O(N)
  • Approach 3: In-Place
    Like quicksort.
    Time Complexity: O(N), Space Complexity: O(1)

代码

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class Solution(object):
    def sortArrayByParity(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        
        ### Sort
        ### Time Complexity: O(NlogN), Space Complexity: O(N)
        # A.sort(key = lambda x: x % 2)
        # return A
        
        ### Two Pass
        ### Time Complexity: O(N), Space Complexity: O(N)
        # return ([x for x in A if x % 2 == 0] + [x for x in A if x % 2 == 1])
        
        ### In-Place
        ### Time Complexity: O(N), Space Complexity: O(1)
        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