leetcode-【hard】4. Median of Two Sorted Arrays

题目

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, 3]
nums2 = [2]

The median is 2.0

 

Example 2:

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

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

答案

看到题目要求O(log (m+n))，知道应该用二分，但是没有想到具体实施的办法，在网上搜了答案，看懂了完成的代码，这道题刷新了我对二分的看法，神一样的存在。

代码

#define MIN(a,b) ((a) > (b) ? (b) : (a))


class Solution {
public:
    int findMedNum(vector<int>::iterator nums1Beg,int nums1Num,vector<int>::iterator nums2Beg,int nums2Num,int med)
    {
        if(nums1Num == 0 )
        {
            return *(nums2Beg + med - 1);
        }
        if(nums1Num > nums2Num)
        {
            return findMedNum(nums2Beg,nums2Num,nums1Beg,nums1Num,med);
        }
        
        if(med == 1)
        {
            return MIN(*nums1Beg,*nums2Beg);
        }
        
        int nums1BegMed = MIN(med / 2, nums1Num);
        int nums2BegMed = med - nums1BegMed;
        
        if(*(nums1Beg + nums1BegMed - 1) == *(nums2Beg + nums2BegMed - 1))
        {
            return *(nums1Beg + nums1BegMed - 1);
        }
        
        if(*(nums1Beg + nums1BegMed - 1) < *(nums2Beg + nums2BegMed - 1))
        {
            return findMedNum(nums1Beg + nums1BegMed,nums1Num - nums1BegMed,nums2Beg,nums2BegMed,med - nums1BegMed);
        }
        
        if(*(nums1Beg + nums1BegMed - 1) > *(nums2Beg + nums2BegMed - 1))
        {
            return findMedNum(nums1Beg,nums1BegMed,nums2Beg + nums2BegMed,nums2Num - nums2BegMed,med - nums2BegMed);
        }
        
        return 0;
    }
    
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        if(nums1.size() == 0 && nums2.size() == 0)
        {
            return 0;
        }
        int totalLen = nums1.size() + nums2.size();
        
        if(totalLen & 0x1)
        {
            return findMedNum(nums1.begin(),nums1.size(),nums2.begin(),nums2.size(),(totalLen / 2) + 1);
        }
        else
        {
            int pre = findMedNum(nums1.begin(),nums1.size(),nums2.begin(),nums2.size(),totalLen / 2);
            int post = findMedNum(nums1.begin(),nums1.size(),nums2.begin(),nums2.size(),(totalLen / 2) + 1);
            return (double)(pre + post) / 2.0;
        }
        
        return 0;
    }
};