LeetCode | 4. Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

Description

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: (Java)

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int index1 = 0, index2 = 0, medianIndex, count = 0;
        medianIndex = (nums1.length + nums2.length) / 2;
        int[] arr = new int[medianIndex + 1];
        while (index1 < nums1.length && index2 < nums2.length) {
            if (nums1[index1] <= nums2[index2]) {
                arr[count] = nums1[index1];
                index1++;
            } else {
                arr[count] = nums2[index2];
                index2++;
            }
            count++;
            if (count > medianIndex)
                break;
        }
        if (count <= medianIndex) {
            if (index1 == nums1.length) {
                while (index2 < nums2.length) {
                    arr[count] = nums2[index2];
                    index2++;
                    count++;
                    if (count > medianIndex)
                        break;
                }
            } else {
                while (index1 < nums1.length) {
                    arr[count] = nums1[index1];
                    index1++;
                    count++;
                    if (count > medianIndex)
                        break;
                }
            }
        }
        if ((nums1.length + nums2.length) % 2 == 0)
            return (arr[medianIndex-1] + arr[medianIndex])*1.0 / 2;
        else
            return arr[medianIndex];
    }
}

思路

本题的代码还可以再稍微精简一下，不过这样也比较简单易懂吧，先得到中位数的索引位置，然后从两个数组中，依次把较小的数往新的数组填充，一直到中位数的位置，那么新数组的最后一个或者最后两个数的平均值即为中位数。