922. Sort Array By Parity II

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
class Solution:
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        odd = []
        even = []
        for i in A:
            if i%2==1:
                odd.append(i)
            else:
                even.append(i)
        pos = 0
        for i in range(len(odd)):
            A[pos] = even[i]
            A[pos+1] = odd[i]
            pos +=2
        return A
posted @ 2018-10-31 11:46  bernieloveslife  阅读(151)  评论(0编辑  收藏  举报