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)).

Solution: 求两个有序数组的中位数，采用分治法，每次缩小搜索空间，分治法的基准点选择两个数组中位数。 该问题要注意m+n分别为奇偶的情况。

class Solution {public:    int findk(vector<int>& a, vector<int>& b, int k){        int n = a.size();        int m = b.size();        if (n <= 0) return b[k-1];          if (m <= 0) return a[k-1];          if (k <= 1) return min(a[0], b[0]);           if (b[m/2] >= a[n/2])          {              if ((n/2 + 1 + m/2) >= k) {                   vector<int> b2(b.begin(), b.begin() + m/2);                  return findk(a, b2, k);             }            else{                 vector<int> a2(a.begin() + n/2 + 1, a.end());                 return findk(a2, b, k - (n/2 + 1));              }                           }          else          {              if ((m/2 + 1 + n/2) >= k){                 vector<int> a2(a.begin(), a.begin() + n/2);                 return findk( a2,b, k);            }            else{                   vector<int> b2(b.begin() + m/2 + 1, b.end());                 return findk( a,b2,k - (m/2 + 1));              }        }      }    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {        int m = nums1.size();        int n = nums2.size();        if((m + n ) %2 == 0)        {            return (findk(nums1, nums2,(m + n)/2 ) + findk(nums1, nums2,(m + n)/2 + 1 )) / 2.0;        }        else            return findk(nums1, nums2,(m + n)/2 + 1);    }};