leetcode 04 Median of Two Sorted Arrays

n1 为 num1的 len
n2 为 num2的 len
故中间的数应该是 k = (n1 + n2 + 1) / 2
二分 num1中位置 m1 , 故 num2的位置为m2
必须保证 nums1[m1-1] <= nums2[m2]
且nums1[m1] >= nums2[m2-1]

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        if(nums1.size() > nums2.size()) 
            return findMedianSortedArrays(nums2, nums1);
        int n1 = nums1.size(), n2 = nums2.size();
        int l = 0, r = n1-1;
        const int k = (n1 + n2 + 1) / 2;
        while(l <= r) {
            int m1 = (l + r) /2;
            int m2 = k - m1;
            if(nums1[m1] <= nums2[m2-1])
                l = m1 + 1;
            else 
                r = m1 - 1;
        }
        int m1 = l;
        int m2 = k - l;
        int c1 = max( m1<=0 ? INT_MIN : nums1[m1-1],
                      m2<=0 ? INT_MIN : nums2[m2-1]);
        int c2 = min( m1>=n1 ? INT_MAX : nums1[m1],
                      m2>=n2 ? INT_MAX : nums2[m2] );
        if((n1 + n2) & 1) 
            return c1;
        return 0.5 * (c1 + c2);
    }
};