LeetCode 4. Median of Two Sorted Arrays 两个排序数组的中位数

4. Median of Two Sorted Arrays

Difficulty: Hard

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)).

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

Language: Java

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if (nums1 == null || nums2 == null) {
            return 0;
        }
        int total = nums1.length + nums2.length;
        if (total % 2 == 0) {
            return (findKth(nums1, nums2, 0, 0, total / 2) + findKth(nums1, nums2, 0, 0, total / 2 + 1)) / 2.0;
        } else {
            return findKth(nums1, nums2, 0, 0, total / 2 + 1);
        }
    }
    
    private int findKth(int[] nums1, int[] nums2, int start1, int start2, int k) {
        if (start1 >= nums1.length) {
            return nums2[start2 + k - 1];
        }
        if (start2 >= nums2.length) {
            return nums1[start1 + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[start1], nums2[start2]);
        }
        int val1 = start1 + k / 2 - 1 < nums1.length ? nums1[start1 + k / 2 - 1] : Integer.MAX_VALUE;
        int val2 = start2 + k / 2 - 1 < nums2.length ? nums2[start2 + k / 2 - 1] : Integer.MAX_VALUE;
        if (val1 < val2) {
            return findKth(nums1, nums2, start1 + k / 2, start2, k - k / 2);
        } else {
            return findKth(nums1, nums2, start1, start2 + k / 2, k - k / 2);
        }
    }
}