4. Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

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Pick One

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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

  int findkth(int a[],int aSize, int b[],int bSize,int k) //求解第k个小的数
    {

        int aPos,bPos;
        if(aSize>bSize)//保证a始终是较短序列
        {
            return findkth(b,bSize,a,aSize,k);
        }
        if(aSize==0)  //如果序列a空了，则直接返回b串的第k个数
            return b[k-1];
        if(k==1)
            return a[0]<b[0]?a[0]:b[0];
        aPos=k/2<aSize?k/2:aSize;
        bPos=k-aPos;
        if(a[aPos-1]==b[bPos-1])
        {
            return a[aPos-1];
        }
        else if(a[aPos-1]<b[bPos-1])
        {
            return findkth(a+aPos,aSize-aPos,b,bSize,k-aPos);
        }
        else
        {
            return findkth(a,aSize,b+bPos,bSize-bPos,k-bPos);
        }

    }

    double findMedianSortedArrays(int* nums1,int nums1Size,int* nums2,int nums2Size)
    {
        int num=nums1Size+nums2Size;
        if(num%2==0)
        {
            return (findkth(nums1,nums1Size,nums2,nums2Size,num/2)+findkth(nums1,nums1Size,nums2,nums2Size,num/2+1))/2.0;
        }
        else
        {
            return findkth(nums1,nums1Size,nums2,nums2Size,num/2+1)*1.0;
        }
    }