Median of Two Sorted Arrays

There are two sorted arrays
nums1 and nums2 of size m and n
respectively.

find the median of the two
sorted arrays. The overall run 
time complexity should be 0(log (m+n)).

you may assume nums1 and nums2
cannot be both empty.

Example 1:
nums1 = [1, 3]
nums2 = [2]
the median is 2.0

Example 2:
nums1= [1, 2]
nums2= [3, 4]
The median is (2+3)/2= 2.5

Solution of this Program:

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) { // to ensure m<=n
            int[] temp = A; A = B; B = temp;
            int tmp = m; m = n; n = tmp;
        }
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j-1] > A[i]){
                iMin = i + 1; // i is too small
            }
            else if (i > iMin && A[i-1] > B[j]) {
                iMax = i - 1; // i is too big
            }
            else { // i is perfect
                int maxLeft = 0;
                if (i == 0) { maxLeft = B[j-1]; }
                else if (j == 0) { maxLeft = A[i-1]; }
                else { maxLeft = Math.max(A[i-1], B[j-1]); }
                if ( (m + n) % 2 == 1 ) { return maxLeft; }

                int minRight = 0;
                if (i == m) { minRight = B[j]; }
                else if (j == n) { minRight = A[i]; }
                else { minRight = Math.min(B[j], A[i]); }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
}

There are other articles are available which gives you detailed information about the array like what is array?, How to declare, initialize, instantiate array. Also, there are some array-related programs are available like the 3sums.

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