LeetCode：4. Median of Two Sorted Arrays(找出两个有序数组的中间数)

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

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方法1：这个比较常见的一种方法；先排序，再进行计算

package leetcode;

import org.junit.Test;


/**
 * @author zhangyu
 * @version V1.0
 * @ClassName: MergeSortArray
 * @Description: 这个题目在leetcode被编为hard类型，我这种做法居然还有这么高的效率，有点不太相信
 * Runtime: 28 ms, faster than 86.93% of Java online submissions for Median of Two Sorted Arrays.
 * @date 2018/12/7 15:12
 **/

public class MergeSortArray {
    @Test
    public void fun() {
        int[] nums1 = {1, 3};
        int[] nums2 = {2, 4};
        double media = getMedian(nums1, nums2);
        // System.out.println(Arrays.toString(newArr));
        System.out.println(media);
    }

    private double getMedian(int[] nums1, int[] nums2) {
        double media = 0;
        int[] newArr = mergeSortedArray(nums1, nums2);
        if (newArr.length % 2 == 0) {
            int length = newArr.length;
            media = (newArr[(length - 1) / 2] + newArr[(length - 1) / 2 + 1]) / 2.0;
        } else {
            media = newArr[(newArr.length - 1) / 2];
        }
        return media;
    }

    public int[] mergeSortedArray(int[] nums1, int[] nums2) {
        int nums1Length = nums1.length;
        int nums2Length = nums2.length;
        int[] newArr = new int[nums1Length + nums2Length];
        int i = 0;
        int j = 0;
        int index = 0;
        while (i < nums1Length && j < nums2Length) {
            if (nums1[i] < nums2[j]) {
                newArr[index++] = nums1[i];
                i++;
            } else {
                newArr[index++] = nums2[j];
                j++;
            }
        }
        while (i < nums1Length) {
            newArr[index++] = nums1[i];
            i++;
        }
        while (j < nums2Length) {
            newArr[index++] = nums2[j];
            j++;
        }
        return newArr;
    }
}

时间复杂度：O(n+m)

空间复杂度：O(n)