Median of Two Sorted Arrays --leetcode java

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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)).——————-

public class FindMedianSortedArrays {    public double findMedianSortedArrays(int[] nums1, int[] nums2) {        double median = -1;        int length_sum = nums1.length + nums2.length;        int[] nums3 = merge(nums1, nums2, length_sum);        if (length_sum % 2 == 1) {            median = nums3[(int) Math.floor(length_sum / 2)];        } else {            //必须强制类型转换，否则结果转化为整形            median = (double)(nums3[length_sum / 2] + nums3[length_sum / 2 - 1]) /(double) 2;        }        return median;    }    public int[] merge(int[] nums1, int[] nums2, int length_sum) {        if (nums1.length == 0)            return nums2;        if (nums2.length == 0)            return nums1;        int[] nums3 = new int[length_sum];        //new一个新的数组，在原来数组的最后多加一个哨兵，避免越界        int[] a=new int[nums1.length+1];        int[] b=new int[nums2.length+1];        for(int i=0;i<nums1.length;i++){            a[i]=nums1[i];        }        for(int j=0;j<nums2.length;j++){            b[j]=nums2[j];        }        //设置一个哨兵牌，避免数组越界        a[nums1.length]=Integer.MAX_VALUE;        b[nums2.length]=Integer.MAX_VALUE;        int i = 0, j = 0;        for (int m = 0; m < length_sum; m++) {            if (a[i] < b[j]) {                nums3[m] = a[i];                i = i + 1;            } else {                nums3[m] = b[j];                j = j + 1;            }        }        return nums3;    }}

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