Sort Array By Parity II
Description
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 * 104
nums.length
is even.- Half of the integers in
nums
are even. 0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
Solution(javascript)
/**
* @param {number[]} nums
* @return {number[]}
*/
var sortArrayByParityII = function(nums) {
let even = 0
let odd = 1
while (even < nums.length && odd < nums.length){
if (nums[even] % 2 === 0) {
even += 2
} else if (nums[odd] % 2 === 1) {
odd += 2
} else {
let temp = nums[odd]
nums[odd] = nums[even]
nums[even] = temp
}
}
return nums
};