The median exercise

Given two ordered arrays of size n, and m.

Please find the median of the two ordered array, and requires time complexity of O (log (m + n) )
Example:
nums1 = [. 1, 2]
nums2 = [. 3,. 4]

The median is (2 + 3) / 2 = 2.5
Analysis:
introducing a third array, the two arrays in an orderly manner into the third array, the third intermediate value to find the array, if third array the number of elements is an even number, the intermediate value between the two values, if it is odd, then to an intermediate value.
Code:

public class median {
    public double findMedianSortedArrays(int[] array1, int[] array2) {
        int i=0,j=0,k=0;
        int[] array3=new int[array1.length+array2.length];         //引入第三个数组
        while(i<array1.length&&j<array2.length){
            if(array1[i]<=array2[j]){
                array3[k++]=array1[i++];
            }
            else if(array1[i]>=array2[j]){
                array3[k++]=array2[j++];
            }

        }
        while(i<array1.length){   //只有一个数组中有元素
            array3[k++]=array1[i++];
        }
        while(j<array2.length){
            array3[k++]=array2[j++];
        }
        //若数组中的元素为偶数个
        if(array3.length%2==0){
            double result=(array3[array3.length/2]+array3[(array3.length-1)/2])/2.0;
            return result;
        }
        double result=array3[array3.length/2];
        return result;
    }

    public static void main(String[] args) {
     int[] array1={1,2,4,5,7,9};
     int[] array2={1,4,5,7,8};
     median m=new median();
        System.out.println( "中间值为：" +m.findMedianSortedArrays(array1,array2));
    }
}