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