LeetCode – Median of Two Sorted Arrays Java

LeetCode Problem:

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)).

Keys to solve this problem

This problem can be converted to the problem of finding kth element, k is (A’s length + B’ Length)/2.

If any of the two arrays is empty, then the kth element is the non-empty array’s kth element.
If k == 0, the kth element is the first element of A or B.

For normal cases(all other cases), we need to move the pointer at the pace of half of an array length.

public static double findMedianSortedArrays(int A[], int B[]) {int m = A.length;int n = B.length; if ((m + n) % 2 != 0) // oddreturn (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);else { // evenreturn (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1) + findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;}} public static int findKth(int A[], int B[], int k, int aStart, int aEnd, int bStart, int bEnd) { int aLen = aEnd - aStart + 1;int bLen = bEnd - bStart + 1; // Handle special casesif (aLen == 0)return B[bStart + k];if (bLen == 0)return A[aStart + k];if (k == 0)return A[aStart] < B[bStart] ? A[aStart] : B[bStart]; int aMid = aLen * k / (aLen + bLen); // a's middle countint bMid = k - aMid - 1; // b's middle count // make aMid and bMid to be array indexaMid = aMid + aStart;bMid = bMid + bStart; if (A[aMid] > B[bMid]) {k = k - (bMid - bStart + 1);aEnd = aMid;bStart = bMid + 1;} else {k = k - (aMid - aStart + 1);bEnd = bMid;aStart = aMid + 1;} return findKth(A, B, k, aStart, aEnd, bStart, bEnd);}