算法描述：

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)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]nums2 = [2]The median is 2.0

Example 2:

nums1 = [1, 2]nums2 = [3, 4]The median is (2 + 3)/2 = 2.5

解题思路：利用二分法将两个数组分别分割，当数组分割结果的所有左边的值小于右边的值，则表示找到对应的值。

double findMedianSortedArrays(vector
     
      & nums1, vector
      
       & nums2) {        if(nums1.size() > nums2.size()) return findMedianSortedArrays(nums2, nums1);        int m = nums1.size();        int n = nums2.size();                int left = 0;        int right = m;        while(left <= right){            int partitionX = (left + right) /2;            int partitionY = (m+n+1)/2 - partitionX;                        int maxLeftX = (partitionX==0) ? INT_MIN : nums1[partitionX-1];            int minRightX = (partitionX == m) ? INT_MAX : nums1[partitionX];                        int maxLeftY = (partitionY ==0) ? INT_MIN : nums2[partitionY-1];            int minRightY = (partitionY == n) ? INT_MAX : nums2[partitionY];                        if(maxLeftX <= minRightY && maxLeftY <= minRightX){                if((m+n)%2==0)                    return (double)(min(minRightY,minRightX)+max(maxLeftX, maxLeftY))/2;                else                    return (double) max(maxLeftX, maxLeftY);            } else if(maxLeftX > minRightY)                right = partitionX - 1;            else                left = partitionX + 1;        }        return -1;    }