Leetcode - 922 Sorting By Parity 2

Question

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.

I've written this code.

public static void main(String[] args) {
        int[] nums = {1,3,5,7,2,4,6,8};
        int[] res = new int[nums.length]; 
        int a=0;
        int b=0;
        int c=1;
        while (a < nums.length && b < nums.length+1 && c < nums.length+1){
            if (nums[a]%2 == 0){
                res[b] = nums[a];
                a++;
                b += 2;
            }
            if (nums[a]%2 != 0){
                res[c] = nums[a];
                a++; 
                c += 2;
            }
        }
        for (int i : res){
            System.out.println(res[i]);
        }
        System.out.println();
    }

While executing the {1,3,5,7,2,4,6,8} in even spaces it fills '0' but, while executing the {2,4,6,8,10,1,3,5,7,9} it gives correct answer. Can someone please help me here?

Answer 1

Because it in the first case, your code puts the odd numbers in the res array, and then stops because of the c < nums.length+1 condition. You only need a < nums.length as while condition.