Algorithm
Problem Name: 922. Sort Array By Parity II
Problem Link: https://leetcode.com/problems/sort-array-by-parity-ii/
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 * 104nums.lengthis even.- Half of the integers in
numsare even. 0 <= nums[i] <= 1000
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int[] sortArrayByParityII(int[] A) {
int start = 1;
for (int i = 0; i < A.length; i += 2) {
if (A[i] % 2 == 1) {
while (A[start] % 2 != 0) {
start += 2;
}
int temp = A[i];
A[i] = A[start];
A[start] = temp;
}
}
return A;
}
}
Input
nums = [4,2,5,7]
Output
[4,5,2,7]
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const sortArrayByParityII = function(A) {
const res = []
const odd = []
const even = []
for(let i = 0, len = A.length; i < len; i++) {
if(A[i] % 2 === 0) even.push(A[i])
else odd.push(A[i])
}
for(let i = 0, len = odd.length; i < len; i++) {
res.push(even[i], odd[i]>
}
return res
};
Input
nums = [4,2,5,7]
Output
[4,5,2,7]
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def sortArrayByParityII(self, A):
even, odd = [a for a in A if not a % 2], [a for a in A if a % 2]
return [even.pop() if not i % 2 else odd.pop() for i in range(len(A))]
Input
nums = [2,3]
Output
[2,3]
#4 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0922_SortArrayByParityII
{
public int[] SortArrayByParityII(int[] A)
{
var oddIndex = 1;
for (int evenIndex = 0; evenIndex < A.Length; evenIndex += 2)
{
if (A[evenIndex] % 2 == 1)
{
while (A[oddIndex] % 2 == 1)
oddIndex += 2;
var temp = A[evenIndex];
A[evenIndex] = A[oddIndex];
A[oddIndex] = temp;
}
}
return A;
}
}
}
Input
nums = [2,3]
Output
[2,3]