【leetCode】4. 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 O(log (m+n)).

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

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int n1 = nums1.length;
        int n2 = nums2.length;
        if ((n1 + n2) % 2 == 1) {
            return findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2 + 1);
        } else {
            return (findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2) + findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2 + 1))
                    / 2;
        }
    }

    public double findNumK(int[] nums1, int n1, int[] nums2, int n2, int k) {
        if (n1 > n2) {
            return findNumK(nums2, n2, nums1, n1, k);
        }
        if (n1 == 0) {
            return nums2[k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[0], nums2[0]);
        }
        int temp1 = Math.min(k / 2, n1);
        int temp2 = k - temp1;
        if (nums1[temp1 - 1] < nums2[temp2 - 1]) {
            return findNumK(subArray(nums1, temp1), n1 - temp1, nums2, n2, k - temp1);
        } else if (nums1[temp1 - 1] > nums2[temp2 - 1]) {
            return findNumK(nums1, n1, subArray(nums2, temp2), n2 - temp2, k - temp2);
        } else {
            return nums1[temp1 - 1];
        }
    }

    public int[] subArray(int[] a, int b) {
        int[] temp = new int[a.length - b];
        for (int i = 0; i < a.length - b; i++) {
            temp[i] = a[b + i];
        }
        return temp;
    }
}