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?
SOLUTION:
class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
n = len(nums)
odds = []
evens = []
for i in range(n - 1, -1, -1):
if nums[i] & 1:
odds.append(nums[i])
else:
evens.append(nums[i])
for i in range(n):
if i & 1:
nums[i] = odds.pop()
else:
nums[i] = evens.pop()
return nums
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