LeetCode 4. 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)).

两个有序数组，找中位数。

public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1 = nums1.length;
        int len2 = nums2.length;
        if(len1==0){
            if(len2%2==1){
                return (double)nums2[len2/2];
            }
            else return ((double)nums2[(len2-1)/2]+nums2[len2/2])/2;
        }
        if(len2==0){
            if(len1%2==1){
                return (double)nums1[len1/2];
            }
            else return ((double)nums1[(len1-1)/2]+nums1[len1/2])/2;
        }
        int mid = (len1+len2-1)/2;
        int midval1=0;
        int midval2=0;
        int i=0,j=0,k=0;
        for(;i<mid;i++){
            if(j==len1){
                k++;
            }
            else if(k==len2||nums1[j]<=nums2[k]){
                j++;
            }
            else{
                k++;
            }
        }
        if(j==len1){
            midval1 = nums2[k];
            midval2 = nums2[k+1];
        }
        else if(k==len2){
            midval1 = nums1[j];
            midval2 = nums1[j+1];
        }
        else{
            midval1 = Math.min(nums1[j],nums2[k]);
            int m=(nums1[j]<=nums2[k])?j++:k++;
            if(j==len1) midval2=nums2[k];
            else if(k==len2) midval2=nums1[j];
            else midval2 = Math.min(nums1[j],nums2[k]);
        }
        if((len1+len2)%2==1) {
            return (double)midval1;
        }
        else return ((double)midval1+midval2)/2;
    }
}

这个代码写的有点冗长，勉强看吧。。 思路就是对两个数组逐个比较计数到mid，如果两个数组长度和是偶数就还要看mid+1