【LeetCode & 剑指offer刷题】查找与排序题4：Median of Two Sorted Arrays

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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

---

C++

 

//方法一：合并，然后计算中值

//不过时间复杂度为O（m+n），不满足题意

class Solution

{

public:

    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)

    {

        int i=0,j=0,k=0;

        int m = nums1.size();

        int n = nums2.size();

        int len = m+n;

        vector<int> temp(len);

        while(i<m&&j<n) //合并，耗时O(m+n)

        {

            temp[k++] = (nums1[i]<nums2[j])? nums1[i++]:nums2[j++];

        }

        while(i<m)

        {

            temp[k++] = nums1[i++];

        }

        while(j<n)

        {

            temp[k++] = nums2[j++];

        }

      

        double mid; //注意这里类型为浮点数，因为可能为小数

        if(len&1)

            mid = temp[(len-1)/2];//长度为奇数时，中值为中间数

        else

            mid = (temp[(len-1)/2]+temp[len/2])/2.0;//长度为偶数时，中值为中间两数平均数

        return mid;

      

      

    }

};

 

//方法二：递归，利用二分查找

//没看懂？？

class Solution

{

public:

    int getkth(vector<int>& s, int m, vector<int>& l, int n, int k)

    {

        //让m<=n

        if(m>n) return getkth(l,n,s,m,k);

        if(m==0) return l[k-1];

        if(k == 1) return min(s[0], l[0]);

       

        int i = min(m, k/2),j = min(n,k/2);

        if(s[i-1] > l[j-1]) return getkth(s,m,, n-j, k-j);

        else return getkth(s+i, m-i, l, n, k-i);

        return 0;

    }

       

    double findMedianSortedArrays(vector<int>& a, vector<int>& b)

    {

        int m = a.size();

        int n = b.size();

        int l = (m+n+1) >> 1;

        int r = (m+n+2) >> 1;

        return (getkth(a,m,b,n,l) + getkth(a,m,b,n,r))/2.0;

              

    }

};