求两个数组的中位数

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)).

Example 1:
nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

Example 2:
nums1 = [1, 3]
nums2 = [2]

The median is 2.0

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
         int len1 = nums1.size();
        int len2 = nums2.size();
        int len = len1+len2;
          int c = 0;
        if(len%2)
        {
            c = (len+1)/2;
        }else
        {
            c= len/2;
        }
        int cen=0;
        int i=0;
        int j=0;
        int n =1;
        int c1=0;
        int c2=0;
        bool nnn = true;
        for(;i<len1&&j<len2;)
        {

            if(nums1[i]<=nums2[j])
            {

                if(n==c)
                {
                    c1=nums1[i];
                }
                if(n==c+1)
                {
                    c2=nums1[i];
                    nnn = false;
                    break;
                }
                i++;
                n++;
                continue;
            }else
            {
                if(n==c)
                {
                    c1=nums2[j];
                }
                if(n==c+1)
                {
                    c2=nums2[j];
                    nnn==false;
                    break;
                }
                j++;
                n++;
                continue;
            }
        }
        for(;i<len1;i++)
        {
            if(n==c)
            {
                c1=nums1[i];
            }
            if(n==c+1&&nnn)
            {
                c2=nums1[i];
                break;
            }
            n++;
        }
        for(;j<len2;j++)
        {

            if(n==c)
            {
                c1=nums2[j];
            }
            if(n==c+1&&nnn)
            {
                c2=nums2[j];
                break;
            }
            n++;
        }
        if(len%2)
        {
            return c1;
        }else
        {
            return (c1+c2)/2.0;
        }
    }
};