Description
Given an integer array nums
, move all the even integers at the beginning of the array followed by all the odd integers.
Return any array that satisfies this condition.
Example 1:
Input: nums = [3,1,2,4] Output: [2,4,3,1] Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Example 2:
Input: nums = [0] Output: [0]
Constraints:
1 <= nums.length <= 5000
0 <= nums[i] <= 5000
Solution
Python3
class Solution:
def sortArrayByParity(self, nums: List[int]) -> List[int]:
i = 0
for j, x in enumerate(nums):
if x % 2 == 0:
nums[i], nums[j] = nums[j], nums[i]
i += 1
return nums
C++
class Solution {
public:
vector<int> sortArrayByParity(vector<int> &A) {
for (int i = 0, j = 0; j < A.size(); j++)
if (A[j] % 2 == 0) swap(A[i++], A[j]);
return A;
}
};