Median of Two Sorted Arrays

There are two sorted arrays A and B 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)).

思路：Ref: http://blog.csdn.net/zxzxy1988/article/details/8587244

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.

class Solution {public:    double findMedianSortedArrays(int A[], int m, int B[], int n) {        int total = m + n;        if (total & 1)  {            return findKth(A, m, B, n, total / 2 + 1); // 如果total是奇数，中位数就是total/2+1位置上的数，即中间的数        }        else {            return (findKth(A, m, B, n, total / 2) + findKth(A, m, B, n, total / 2 + 1))/2; //如果total是偶数，中间两个数的平均数为中间数        }    }        // 寻找第k个数, k从1开始    double findKth(int* A, int m, int* B, int n, int k) {        // 总是假设参数中m要比n小，所以后面也是通过m为0判断是否将A删除完了        if (m > n) {            return findKth(B, n, A, m, k);        }                if (m == 0) {            return B[k - 1]; // 如果A已经删除完了，那么直接返回B的第k-1个即为所寻找的第k个元素（k从1开始计数）        }                // 注意m==0要放k==1前面，防止出现A[0]不存在的情况会报错；        if (k == 1) {            return min(A[0], B[0]);        }                int pa = min(k / 2, m); //  防止出现m小于k/2的情况        int pb = k - pa; // 正常情况也是k/2        if (A[pa - 1] < B[pb - 1]) {            return findKth(A + pa, m - pa, B, n, k - pa); // 删除A的前pa个，同时寻找第k-pa个        }        else if (A[pa - 1] > B[pb - 1]) {            return findKth(A, m, B + pb, n - pb, k - pb); // 删除B的前pb个，同时寻找第k-pb个        }        else {            return A[pa - 1];        }    }};