LeetCode-4-Median of Two Sorted Arrays

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<int>& nums1, vector<int>& 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;
    }