4.Median of Two Sorted Arrays Leetcode Python

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)).

这题的一种O(m+n)的解法是mergesort的一步，依次比较两者的大小，将得到的解放到新的item里面。

最后再取出median

class Solution:
    # @return a float
    def findMedianSortedArrays(self, A, B):
        item=[]
        while len(A)>0 and len(B)>0:
            item.append(A.pop(0) if A[0]<=B[0] else B[0])
        while len(A)>0:
            item.append(A.pop(0))
        while len(B)>0:
            item.append(B.pop(0))
        if len(item)%2==1:
            return item[len(item)/2]
        else: 
            return (item[len(item)/2]+item[len(item)/2-1])/2
            
        

另外一种解法是来自
http://blog.youkuaiyun.com/zxzxy1988/article/details/8587244

主题思想是每次比较A和B的各一半 

如果A[half]<B[half]  中值会在 A剩下的一半和B中产生

反之 在A和B剩下的一半产生。

如果相等就刚刚好返回A[half]

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:
    # @return a float
    def findk(self,A,m,B,n,k):
        if m>n:
            return self.findk(B,n,A,m,k)
        if m==0:
            return B[k-1]
        if k==1:
            return min(A[0],B[0])
        pa=min(k/2,m)
        pb=k-pa
        if A[pa-1]<B[pb-1]:
            return self.findk(A[pa:],m-pa,B,n,k-pa)
        elif A[pa-1]>B[pb-1]:
            return self.findk(A,m,B[pb:],n-pb,k-pb)
        else:
            return A[pa-1]
        
    def solve(self,A,m,B,n):
        m=len(A)
        n=len(B)
        total=m+n
        if total%2==1:
            return self.findk(A,m,B,n,total/2+1)
        else:
            return (self.findk(A,m,B,n,total/2+1)+self.findk(A,m,B,n,total/2))/2.0
            
    def findMedianSortedArrays(self, A, B):
        m=len(A)
        n=len(B)
        result=self.solve(A,m,B,n)
        return result