sort-array-by-parity-ii.go

// 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?
//


func sortArrayByParityII(A []int) []int {
    result := make([]int, len(A))
    evenP := 0
    oddP := 1
    for i:=0; i<len(A); i++ {
        current := A[i]
        if current % 2 == 0 {
            result[evenP] = current
            evenP += 2
        } else {
            result[oddP] = current
            oddP += 2
        }
    }
    return result
}