[leedcode 04]Median of Two Sorted Arrays

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 Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        //此题利用二分法思想。题眼：看题目要求时间复杂度为O（log（m+n））
        //通过每个数组的中位数将数组分成四个部分。通过判断k值和两个数组数组的中位数，按场景，不断舍弃每部分。此题需要注意下标的处理！！一定           要心细
        int n1=nums1.length;
        int n2=nums2.length;
        if((n1+n2)%2==0){
            return (find(nums1,0,n1,nums2,0,n2,(n1+n2)/2)+find(nums1,0,n1,nums2,0,n2,(n1+n2)/2-1))/2.0;
            
        }else{
            return find(nums1,0,n1,nums2,0,n2,(n1+n2)/2);
        }
    }
    public double find(int[] nums1,int start1,int end1,int[] nums2,int start2,int end2,int k){
        if(start1>=end1){
            return nums2[k+start2];
        }else if(start2>=end2){
            return nums1[k+start1];
        }
        int mid1=(start1+end1)/2;
        int mid2=(start2+end2)/2;
        if(k<=0){
            return Math.min(nums1[start1],nums2[start2]);
        }
        if(k>mid1-start1+mid2-start2){
             if(nums1[mid1]>=nums2[mid2]){
                return find(nums1,start1,end1,nums2,mid2+1,end2,k-(mid2-start2)-1);
            }else{
               return find(nums1,mid1+1,end1,nums2,start2,end2,k-(mid1-start1)-1);
            }
        }else{
             if(nums1[mid1]>=nums2[mid2]){
                return find(nums1,start1,mid1,nums2,start2,end2,k);
            }else{
                return find(nums1,start1,end1,nums2,start2,mid2,k);
            }
        }
    }
}