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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
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;
}
public int[] sortArrayByParity(int[] A) {
for (int i = 0, j = 0; j < A.length; j++)
if (A[j] % 2 == 0) {
int tmp = A[i];
A[i++] = A[j];
A[j] = tmp;;
}
return A;
}
class Solution:
def sortArrayByParity(self, A):
beg, end = 0, len(A) - 1
while beg <= end:
if A[beg] % 2 == 0:
beg += 1
else:
A[beg], A[end] = A[end], A[beg]
end -= 1
return A
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