Given an array of integers nums, half of the integers in nums are odd, and the other half are even.
Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.
Return any answer array that satisfies this condition.
Example 1:
Input: nums = [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.
Example 2:
Input: nums = [2,3]
Output: [2,3]
Constraints:
- 2 <= nums.length <= 2 * 104
- nums.length is even.
- Half of the integers in nums are even.
- 0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
#https://leetcode.com/problems/sort-array-by-parity-ii/
class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
no=1
ne=0
alist=[0]*len(nums)
print(alist)
for i in nums:
if i%2==0:
alist[ne]=i
ne+=2
else:
alist[no]=i
no+=2
return alist
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