905. Sort Array By Parity

Posted on In Words count: 844 Reading time ≈ 1 mins.

905. Sort Array By Parity

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
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

Difficulty : Easy

Solution

two pointers

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int l = 0;
        int r = nums.length - 1;
        while (l < r) {
            while (l < r && nums[l] % 2 == 0) {
                l++;
            }
            while (l < r && nums[r] % 2!= 0) {
                r--;
            }
            if (l < r) {
                swap(nums, l ,r);
            }
        }
        return nums;
    }
    private void swap(int[] nums, int l, int r) {
        int tmp = nums[l];
        nums[l] = nums[r];
        nums[r] = tmp;
    }
}