Median of Two Sorted Arrays @LeetCode

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)).

	package leetcode;
	/**
	* Solution: need some reasoning. This problem is extend to find kth number of two Sorted Array.
	* Just copy other's solution, I didn't understand this problems. Will update later.
	*
	*
	* Reference: http://fisherlei.blogspot.com/2012/12/leetcode-median-of-two-sorted-arrays.html
	*
	* @author jeffwan
	* @date Apr 9, 2014
	*/
	public class FindMedianSortedArrays {
	public double findMedianSortedArrays(int A[], int B[]) {
	int len = A.length + B.length;
	if (len % 2 == 0) {
	return (findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)) / 2.0 ;
	} else {
	return findKth(A, 0, B, 0, len / 2 + 1);
	}
	}
	// find kth number of two sorted array
	public int findKth(int[] A, int A_start, int[] B, int B_start, int k){
	// System.out.println("A_start: "+A_start+" B_start: "+B_start+" k: "+k);
	if(A_start >= A.length)
	return B[B_start + k - 1];
	if(B_start >= B.length)
	return A[A_start + k - 1];
	if (k == 1)
	return Math.min(A[A_start], B[B_start]);
	int A_key = A_start + k / 2 - 1 < A.length
	? A[A_start + k / 2 - 1]
	: Integer.MAX_VALUE;
	int B_key = B_start + k / 2 - 1 < B.length
	? B[B_start + k / 2 - 1]
	: Integer.MAX_VALUE;
	if (A_key < B_key) {
	return findKth(A, A_start + k / 2, B, B_start, k - k / 2);
	} else {
	return findKth(A, A_start, B, B_start + k / 2, k - k / 2);
	}
	}
	// O(n) doesn't meet need of Problem
	public double findMedianSortedArrays2(int A[], int B[]) {
	int[] C = new int[A.length + B.length];
	int i = 0;
	int j = 0;
	int index = 0;
	while (i < A.length && j < B.length) {
	if (A[i] < B[j]) {
	C[index] = A[i];
	i++;
	} else {
	C[index] = B[j];
	j++;
	}
	index++;
	}
	while (i < A.length) {
	C[index] = A[i];
	i++;
	index++;
	}
	while (j < B.length) {
	C[index] = B[j];
	j++;
	index++;
	}
	// Caculate median
	int start, end, mid;
	start = 0;
	end = C.length - 1;
	mid = start + (end - start) / 2;
	if (C.length % 2 == 1) {
	return C[mid];
	} else {
	return (double) (C[mid] + C[mid + 1]) / 2;
	}
	}
	}

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