Twitter OA prepare: even sum pairs

Write a function:
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns the number of pairs (P, Q) such that 0 ≤ P < Q < N and (A[P] + A[Q]) is even. The function should return −1 if the number of such pairs exceeds 1,000,000,000.
For example, given array A such that:
A[0] = 2, A[1] = 1, A[2] = 5, A[3] = −6, A[4] = 9
the function should return 4, because there are four pairs that fulfill the above condition, namely (0,3), (1,2), (1,4), (2,4).
Assume that:
N is an integer within the range [0..1,000,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

思路：无非就是扫描一遍记录奇数和偶数各自的个数，比如为M和N，然后就是奇数里面选两个、偶数里面选两个、一奇一偶，答案就是M(M-1)/2 + N(N-1)/2 + MN