LeetCode – Median of Two Sorted Arrays (Java)
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)).
Java Solution
If we see log(n), we should think about using binary something.
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 nonempty 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 to get log(n) time.
public static double findMedianSortedArrays(int A[], int B[]) { int m = A.length; int n = B.length; if ((m + n) % 2 != 0) // odd return (double) findKth(A, B, (m + n) / 2, 0, m  1, 0, n  1); else { // even return (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 cases if (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 count int bMid = k  aMid  1; // b's middle count // make aMid and bMid to be array index aMid = 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); } 
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