leetcode之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)).

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

居然还是一道难度为困难的题，在我看来似乎没有，第一眼看到这题，就想到了，用一个容器存储所有的数据，然后分两种情况找中值（最复杂的）；还是比较好想和理解的，所以我说没有困难难度，但看讨论区里，都是O（1）空间复杂度，O(log（min(m,n)）)时间复杂度。。。

先贴自己写的复杂代码：

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        double res=0.0;
        int len1=nums1.size();
        int len2=nums2.size();
        vector<int>tmp;
        int i=0,j=0;
        while(i<len1&&j<len2){
            if(nums1[i]<nums2[j])
                tmp.push_back(nums1[i++]);
            else
                tmp.push_back(nums2[j++]);
        }
        while(i<len1)
            tmp.push_back(nums1[i++]);
        while(j<len2)
            tmp.push_back(nums2[j++]);
        int mid=(len1+len2)/2;
        if((len1+len2)%2==0){
            res=(double(tmp[mid])+double(tmp[mid-1]))/2;
        }
        else{
            res=double(tmp[mid]);
        }
        return res;
    }
};

看了下讨论区的代码，其实就是不用空间存储，然后用二分法找到中值，所以才是O(log(min(n,m)))；

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int N1 = nums1.size();
        int N2 = nums2.size();
        if (N1 < N2) return findMedianSortedArrays(nums2, nums1); // Make sure A2 is the shorter one.

        int lo = 0, hi = N2 * 2;
        while (lo <= hi) {
            int mid2 = (lo + hi) / 2;   // Try Cut 2 
            int mid1 = N1 + N2 - mid2;  // Calculate Cut 1 accordingly
            
            double L1 = (mid1 == 0) ? INT_MIN : nums1[(mid1-1)/2]; // Get L1, R1, L2, R2 respectively
            double L2 = (mid2 == 0) ? INT_MIN : nums2[(mid2-1)/2];
            double R1 = (mid1 == N1 * 2) ? INT_MAX : nums1[(mid1)/2];
            double R2 = (mid2 == N2 * 2) ? INT_MAX : nums2[(mid2)/2];


            if (L1 > R2) lo = mid2 + 1; // A1's lower half is too big; need to move C1 left (C2 right)
            else if (L2 > R1) hi = mid2 - 1; // A2's lower half too big; need to move C2 left.
            else return (max(L1,L2) + min(R1, R2)) / 2; // Otherwise, that's the right cut.
            }
        return -1;
    }
};