Coding笔记——Median of Two Sorted Arrays

package com.kzl.leetcode.problem_1_50;


/**
 * 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
 */
public class MedianOfTwoSortedArrays {

    public static void main(String[] args) {

        int[] a = new int[]{};
        int[] b = new int[]{2};

        System.out.println(findMedianSortedArrays(a,b));

    }


    public static double findMedianSortedArrays(int[] nums1, int[] nums2) {

        int n1 = nums1.length;
        int n2 = nums2.length;


        int media = (n1 + n2) / 2;
        int[] num = new int[n1 + n2 + 1];

        int i = 0, j = 0;
        int count = 0;

        // 先相继比较nums1和nums2中的数，按小大顺序插入num中
        while (i < n1 && j < n2) {
            if (nums1[i] < nums2[j]) {
                num[count++] = nums1[i++];
            } else {
                num[count++] = nums2[j++];
            }
        }


        // 如果num1遍历完了
        if (i == n1) {
            while ( j < n2) {
                num[count++] = nums2[j++];
            }
        }

        // 如果num2遍历完了
        if (j == n2) {
            while ( i < n1) {
                num[count++] = nums1[i++];
            }
        }

        // 如果两个数组长度和为偶数
        if ((n1 + n2) % 2 == 0) {
            return (num[media-1] + num[media]) / 2.0;
        } else {
            return 1.0 * num[media];
        }

    }
}