Divide and Conquer majority element algorithm
It's not correct. On input [a a b c], the left side yields [a, 2], the right side yields [nil, 1]. They would be merged into [a, 2].

I'll outline a linear time solution. If there exists a value $x$ such that more than $n/2$ of the elements of $A$ have value $x$, then certainly the $n/2$-th order statistic of $A$ must be $x$. The selection problem is solved in linear time using, for example, median of medians; checking that such value does indeed appear at least $n/2$ times is trivially linear time.