Leetcode 4 Median of Two Sorted Arrays

今天发现了leetcode上面一道题，觉得非常经典，记录之。

题目是这样的：给定两个已经排序好的数组（可能为空），找到两者所有元素中第k大的元素。另外一种更加具体的形式是，找到所有元素的中位数。本篇文章我们只讨论更加一般性的问题：如何找到两个数组中第k大的元素？不过，测试是用的两个数组的中位数的题目，Leetcode第4题 Median of Two Sorted Arrays

方案1：假设两个数组总共有n个元素，那么显然我们有用O(n)时间和O(n)空间的方法：用merge sort的思路排序，排序好的数组取出下标为k-1的元素就是我们需要的答案。

这个方法比较容易想到，但是有没有更好的方法呢？

方案2：我们可以发现，现在我们是不需要“排序”这么复杂的操作的，因为我们仅仅需要第k大的元素。我们可以用一个计数器，记录当前已经找到第m大的元素了。同时我们使用两个指针pA和pB，分别指向A和B数组的第一个元素。使用类似于merge sort的原理，如果数组A当前元素小，那么pA++，同时m++。如果数组B当前元素小，那么pB++，同时m++。最终当m等于k的时候，就得到了我们的答案——O(k)时间，O(1)空间。

但是，当k很接近于n的时候，这个方法还是很费时间的。当然，我们可以判断一下，如果k比n/2大的话，我们可以从最大的元素开始找。但是如果我们要找所有元素的中位数呢？时间还是O(n/2)=O(n)的。有没有更好的方案呢？

我们可以考虑从k入手。如果我们每次都能够剔除一个一定在第k大元素之前的元素，那么我们需要进行k次。但是如果每次我们都剔除一半呢？所以用这种类似于二分的思想，我们可以这样考虑：

Assume that the number of elements in A and B are both larger than k/2, and if we compare the k/2-th smallest element in A(i.e. A[k/2-1]) and the k-th smallest element in B(i.e. B[k/2 - 1]), there are three results:

(Becasue k can be odd or even number, so we assume k is even number here for simplicy. The following is also true when k is an odd number.)

A[k/2-1] = B[k/2-1]

A[k/2-1] > B[k/2-1]

A[k/2-1] < B[k/2-1]

if A[k/2-1] < B[k/2-1], that means all the elements from A[0] to A[k/2-1](i.e. the k/2 smallest elements in A) are in the range of k smallest elements in the union of A and B. Or, in the other word, A[k/2 - 1] can never be larger than the k-th smalleset element
in the union of A and B.

Why?

We can use a proof by contradiction. Since A[k/2 - 1] is larger than the k-th smallest element in the union of A and B, then we assume it is the (k+1)-th smallest one. Since it is smaller than B[k/2 - 1], then B[k/2 - 1] should be at least the (k+2)-th smallest
one. So there are at most (k/2-1) elements smaller than A[k/2-1] in A, and at most (k/2 - 1) elements smaller than A[k/2-1] in B.So the total number is k/2+k/2-2, which, no matter when k is odd or even, is surly smaller than k(since A[k/2-1] is the (k+1)-th
smallest element). So A[k/2-1] can never larger than the k-th smallest element in the union of A and B if A[k/2-1]<B[k/2-1];

Since there is such an important conclusion, we can safely drop the first k/2 element in A, which are definitaly smaller than k-th element in the union of A and B. This is also true for the A[k/2-1] > B[k/2-1] condition, which we should drop the elements in
B.

When A[k/2-1] = B[k/2-1], then we have found the k-th smallest element, that is the equal element, we can call it m. There are each (k/2-1) numbers smaller than m in A and B, so m must be the k-th smallest number. So we can call a function recursively, when
A[k/2-1] < B[k/2-1], we drop the elements in A, else we drop the elements in B.

We should also consider the edge case, that is, when should we stop?

1. When A or B is empty, we return B[k-1]( or A[k-1]), respectively;

2. When k is 1(when A and B are both not empty), we return the smaller one of A[0] and B[0]

3. When A[k/2-1] = B[k/2-1], we should return one of them

In the code, we check if m is larger than n to garentee that the we always know the smaller array, for coding simplicy.

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double findKth(int a[], int m, int b[], int n, int k)  

{  

    //always assume that m is equal or smaller than n  

    if (m > n)  

        return findKth(b, n, a, m, k);  

    if (m == 0)  

        return b[k - 1];  

    if (k == 1)  

        return min(a[0], b[0]);  

    //divide k into two parts  

    int pa = min(k / 2, m), pb = k - pa;  

    if (a[pa - 1] < b[pb - 1])  

        return findKth(a + pa, m - pa, b, n, k - pa);  

    else if (a[pa - 1] > b[pb - 1])  

        return findKth(a, m, b + pb, n - pb, k - pb);  

    else  

        return a[pa - 1];  

}  

  

class Solution  

{  

public:  

    double findMedianSortedArrays(int A[], int m, int B[], int n)  

    {  

        int total = m + n;  

        if (total & 0x1)  

            return findKth(A, m, B, n, total / 2 + 1);  

        else  

            return (findKth(A, m, B, n, total / 2)  

                    + findKth(A, m, B, n, total / 2 + 1)) / 2;  

    }  

};