Median of Two Sorted Arrays

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log(m+n)).
O(m + n) solution:

	public static float getMedian(int[] a, int[] b) {
		if (a == null || b == null || (a.length + b.length) == 0) return Float.MIN_VALUE;
		int pa = 0;
		int pb = 0;
		float median = 0;
		while (pa + pb != (a.length + b.length + 1) / 2) {
			int Ai = (pa == a.length) ? Integer.MAX_VALUE : a[pa];  
	        int Bj = (pb == b.length) ? Integer.MAX_VALUE : b[pb]; 
	        
	        if (Ai < Bj) {
	        	median = a[pa];
	        	pa++;
	        } else {
	        	median = b[pb];
	        	pb++;
	        }
		}
		if ((a.length + b.length) % 2 == 1) {
			return median;	
		} else {
			int Ai = (pa == a.length) ? Integer.MAX_VALUE : a[pa];  
	        int Bj = (pb == b.length) ? Integer.MAX_VALUE : b[pb]; 
			float median2 = (Ai < Bj) ? Ai : Bj;			
			return (median + median2) / 2;
		}
	}

O(log(m+n) solution:

    public int findKthSmallest(int[] A, int[] B, int pa, int delta, int k) {
    	int pb = (k - 1) - pa;
		int Ai_1 = ((pa == 0) ? Integer.MIN_VALUE : A[pa-1]);
		int Bj_1 = ((pb == 0) ? Integer.MIN_VALUE : B[pb-1]);
		int Ai   = ((pa == A.length) ? Integer.MAX_VALUE : A[pa]);
		int Bj   = ((pb == B.length) ? Integer.MAX_VALUE : B[pb]);
		
		if (Bj_1 <= Ai && Ai <= Bj) return Ai;
		if (Ai_1 <= Bj && Bj <= Ai) return Bj;
		
		if (Ai > Bj) {
			pa = pa - delta;
			if (((k - 1) - pa) > B.length) {
				pa = k - 1 - B.length;
			}
		} else {
			pa = pa + delta;
			if (pa > Math.min(A.length, k - 1)) {
				pa = Math.min(A.length, k - 1);
			}
		}
		return findKthSmallest(A, B, pa, (delta + 1) / 2, k);	
	}