一、题目描述

给定两个大小分别为 m 和 n 的正序（从小到大）数组 nums1 和 nums2。请你找出并返回这两个正序数组的 中位数 。

示例 1：

输入：nums1 = [1,3], nums2 = [2]
输出：2.00000
解释：合并数组 = [1,2,3] ，中位数 2
示例 2：

输入：nums1 = [1,2], nums2 = [3,4]
输出：2.50000
解释：合并数组 = [1,2,3,4] ，中位数 (2 + 3) / 2 = 2.5
示例 3：

输入：nums1 = [0,0], nums2 = [0,0]
输出：0.00000
示例 4：

输入：nums1 = [], nums2 = [1]
输出：1.00000
示例 5：

输入：nums1 = [2], nums2 = []
输出：2.00000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/median-of-two-sorted-arrays
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

二、完整代码

class Solution {

    public static void main(String[] args) {
        int nums1[] = {0, 0, 0, 0, 0};
        int nums2[] = {-1, 0, 0, 0, 0, 0, 1};

        double median_num = findMedianSortedArrays(nums1, nums2);
        System.out.println("Median Number is ：" + median_num);

    }

    public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
        double sum = 0.00000;
        int num_len1 = nums1.length;
        int num_len2 = nums2.length;
        if (num_len1 > 1000 || num_len2 > 1000) {
            return sum;
        }
        int merge_num[] = new int[num_len1 + num_len2];

        int num1_index = 0;
        int num2_index = 0;
        int merge_index = 0;
        while (num1_index < num_len1 && num2_index < num_len2) {

            if (nums1[num1_index] < nums2[num2_index]) {
                merge_num[merge_index++] = nums1[num1_index++];
            } else if (nums1[num1_index] > nums2[num2_index]) {
                merge_num[merge_index++] = nums2[num2_index++];
            } else {
                merge_num[merge_index++] = nums1[num1_index++];
            }

        }
        if (num1_index == num_len1) {
            while (num2_index < num_len2)
                merge_num[merge_index++] = nums2[num2_index++];
        } else {
            while (num1_index < num_len1)
                merge_num[merge_index++] = nums1[num1_index++];
        }

        if (merge_num.length % 2 == 0) {
            sum = (merge_num[merge_num.length / 2 - 1] + merge_num[merge_num.length / 2]) / 2.00000;
        } else {
            sum = merge_num[(merge_num.length - 1) / 2];
        }
        return sum;
    }
}