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)).

我的时间复杂度为O(n)，其实仅为(m+n)/2

public class Solution {

    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1 = nums1.length, len2 = nums2.length;
    if(len1 == 0 && len2 == 0){
    return 0;
    }    
    int cou = len1 + len2;
    int median = 0;                           
    boolean isEven = false;
    if(cou % 2 == 1){
    median = cou/2+1;    
    }else{
    median = cou/2;
    isEven = true;
    }    
   
int i = 0, j = 0;
while(i+j < median-1){    
if(i < len1 && j < len2){
if(nums1[i] <= nums2[j]){
i++;
}else{
j++;
}
}else if(j < len2){
j++;
}else{
i++;
}
}
    if(isEven){
    if(i < len1 && j < len2){
    int i1 = 0, i2 = 0;
    if(nums1[i] < nums2[j]){
    i1 = nums1[i];
    if(i+1<len1){
    i2 = Math.min(nums1[i+1], nums2[j]);
    }else{
    i2 = nums2[j];
    }
    }else if(nums1[i] > nums2[j]){
    i1 = nums2[j];
    if(j+1 < len2){
    i2 = Math.min(nums1[i], nums2[j+1]);
    }else{
    i2 = nums1[i];
    }
    }else{
    i1 = nums1[i];
    i2 = nums2[j];
    }
    return (i2+i1)/2.0;    
    }else if(i < len1){
    return (nums1[i]+nums1[i+1])/2.0;
    }else{
    return (nums2[j]+nums2[j+1])/2.0;
    }
    }else{
    if(i < len1 && j < len2){
    return Math.min(nums1[i], nums2[j]);
    }else if(i < len1){
    return nums1[i];
    }else{
    return nums2[j];
    }
    }
    }
}