Median_of_Two_Sorted_Arrays（理论支持和算法总结）

可以将这个题目推广到更naive的情况，找两个排序数组中的第K个最大值（第K个最小值）。

1、直接 merge 两个数组，然后求中位数（第K个最大值或者第K个最小值），能过，不过复杂度是 O(n + m)

python

1 class Solution(object):
2     def findMedianSortedArrays(self, nums1, nums2):
3         """
4         :type nums1: List[int]
5         :type nums2: List[int]
6         :rtype: float
7         """
8
9         tmpresult = []
10         n1 = 0
11         n2 = 0
12
13         while n1 < len(nums1) or n2 < len(nums2):
14             if n1 == len(nums1):
15                 tmpresult.append(nums2[n2])
16                 n2 += 1
17                 continue
18             if n2 == len(nums2):
19                 tmpresult.append(nums1[n1])
20                 n1 += 1
21                 continue
22             if nums1[n1] < nums2[n2]:
23                 tmpresult.append(nums1[n1])
24                 n1 += 1
25             else:
26                 tmpresult.append(nums2[n2])
27                 n2 += 1
28
29         if (n1+n2)&1 :
30             return tmpresult[(n1+n2)/2]
31         else:
32             return (tmpresult[(n1+n2)/2 - 1] + tmpresult[(n1+n2)/2])/2.0

2、不用直接merge两个数组，借助merge的思想，用两个指针pa和pb访问两个数组，size记录当前的找到了第几个数。时间复杂度O(k), 但是当k 很接近m+n时，这个方法还是O(m+n)。

python

1         n1 = 0
2         n2 = 0
3         left = -1
4         right = -1
5         midposition = 0
6         leftnum = -1
7         rightnum = -1
8         if ((len(nums1) + len(nums2)) & 1):
9             left = right = (len(nums1) + len(nums2))/2
10         else:
11             left = (len(nums1) + len(nums2))/2 - 1
12             right = (len(nums1) + len(nums2))/2
13
14         if len(nums1) == 0:
15             return (nums2[left] + nums2[right])/2.0
16         if len(nums2) == 0:
17             return (nums1[left] + nums1[right])/2.0
18
19         while n1 < len(nums1) or n2 < len(nums2):
20
21             if n1 == len(nums1):
22                 if midposition == left:
23                     leftnum = nums2[n2]
24                 if midposition == right:
25                     rightnum = nums2[n2]
26                 midposition += 1
27                 if midposition > right:
28                     break
29                 n2 += 1
30                 continue
31
32             if n2 == len(nums2):
33                 if midposition == left:
34                     leftnum = nums1[n1]
35                 if midposition == right:
36                     rightnum = nums1[n1]
37                 midposition += 1
38                 if midposition > right:
39                     break
40                 n1 += 1
41                 continue
42
43             if nums1[n1] <= nums2[n2]:
44
45                 if midposition == left:
46                     leftnum = nums1[n1]
47                 if midposition == right:
48                     rightnum = nums1[n1]
49                 n1 += 1
50                 midposition += 1
51
52             else:
53
54                 if midposition == left:
55                     leftnum = nums2[n2]
56                 if midposition == right:
57                     rightnum = nums2[n2]
58                 n2 += 1
59                 midposition += 1
60
61             if midposition > right:
62                 break
63
64         return (leftnum+rightnum)/2.0

3、二分查找的思想

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?（反证法证明）
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]
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.

1 class Solution(object):
2     def min(self,a,b):
3         if a>b:
4             return b
5         else:
6             return a
7
8     def findKthsortedarray(self,nums1,nums2,k):
9         if len(nums1) > len(nums2):
10             tmp = nums1
11             nums1 = nums2
12             nums2 = tmp
13         if len(nums1) == 0:
14             return nums2[k-1]
15         if k == 1:
16             return min(nums1[0],nums2[0])
17
18         p1 = min(k/2,len(nums1))
19         p2 = k - p1
20
21         if nums1[p1-1] == nums2[p2-1]:
22             return nums1[p1-1]
23         if nums1[p1-1] > nums2[p2-1]:
24             return self.findKthsortedarray(nums1,nums2[p2:],k-p2)
25         if nums1[p1-1] < nums2[p2-1]:
26             return self.findKthsortedarray(nums1[p1:],nums2,k-p1)
27
28     def findMedianSortedArrays(self, nums1, nums2):
29         num1 = len(nums1)
30         num2 = len(nums2)
31
32         if (num1+num2)&1:
33             return self.findKthsortedarray(nums1,nums2,(num1+num2)/2+1)
34         else:
35             return (self.findKthsortedarray(nums1,nums2,(num1+num2)/2) + self.findKthsortedarray(nums1,nums2,(num1+num2)/2+1))/2.0