##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/905.html

# 905. Sort Array By Parity (Easy)

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.


Note:

1. 1 <= A.length <= 5000
2. 0 <= A[i] <= 5000

## Solution 1. STL

// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
sort(begin(A), end(A), [](int a, int b) { return a % 2 < b % 2; });
return A;
}
};


## Solution 2. Two Pointers

// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
for (int i = 0, j = A.size() - 1; i < j; ++i, --j) {
while (i < j && A[i] % 2 == 0) ++i;
while (i < j && A[j] % 2 != 0) --j;
if (i < j) swap(A[i], A[j]);
}
return A;
}
};

• class Solution {
public int[] sortArrayByParity(int[] A) {
if (A == null || A.length < 2)
return A;
int length = A.length;
for (int i = 1; i < length; i++) {
int num = A[i];
if (num % 2 != 0)
continue;
int insertIndex = 0;
for (int j = i - 1; j >= 0; j--) {
if (A[j] % 2 == 0) {
insertIndex = j + 1;
break;
} else
A[j + 1] = A[j];
}
A[insertIndex] = num;
}
return A;
}
}

############

class Solution {
public int[] sortArrayByParity(int[] nums) {
for (int i = 0, j = nums.length - 1; i < j;) {
if (nums[i] % 2 == 1) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
--j;
} else {
++i;
}
}
return nums;
}
}

• // OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
sort(begin(A), end(A), [](int a, int b) { return a % 2 < b % 2; });
return A;
}
};

• class Solution:
def sortArrayByParity(self, nums: List[int]) -> List[int]:
i, j = 0, len(nums) - 1
while i < j:
if nums[i] & 1:
nums[i], nums[j] = nums[j], nums[i]
j -= 1
else:
i += 1
return nums

############

class Solution(object):
def sortArrayByParity(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
return sorted(A, key = lambda x : x % 2)

• func sortArrayByParity(nums []int) []int {
for i, j := 0, len(nums)-1; i < j; {
if nums[i]%2 == 1 {
nums[i], nums[j] = nums[j], nums[i]
j--
} else {
i++
}
}
return nums
}

• /**
* @param {number[]} nums
* @return {number[]}
*/
var sortArrayByParity = function (nums) {
for (let i = 0, j = nums.length - 1; i < j; ) {
if (nums[i] & 1) {
[nums[i], nums[j]] = [nums[j], nums[i]];
--j;
} else {
++i;
}
}
return nums;
};


• impl Solution {
pub fn sort_array_by_parity(mut nums: Vec<i32>) -> Vec<i32> {
let (mut l, mut r) = (0, nums.len() - 1);
while l < r {
while l < r && nums[l] & 1 == 0 {
l += 1;
}
while l < r && nums[r] & 1 == 1 {
r -= 1;
}
nums.swap(l, r);
}
nums
}
}