Question: http://oj.leetcode.com/problems/median-of-two-sorted-arrays/

Question Name: Median of Two Sorted Arrays

This question is quite similar with the nu2011 (DoubleMedian) by Codility. Both of them are the variants of the question: find the kth smallest element in two sorted array.

BTW: LeetCode’ online judge system is much worse than that from Codility. The problem description did not indicate the data type, whether int or double. In addition, what should we do if they are two empty array? What’s worse, the online judge system is quite unstable. The completely same solution gets two different results: one “Timeout” and one “Accepted”.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 | class Solution: # @return a float def findMedianSortedArrays(self, A, B): def findKthItem(A, B, k): # Find the kth smallest element in the two sorted arrays A and B. # Swap the arrays A and B if needed. Make sure that A is the non-bigger one. # Otherwise, the stepsB might be out of range. For example, # A = [2,3,4,5,6,7,8,9,10], B = [1], k = 9, the stepsA would be 4, and stepsB # would be 3, if not swapped. # # ATTENTION: the stepsA might be -1 if len(A) is zero. But it is harmless # in the next if-elif-else block. if len(A) > len(B): A, B = B, A # stepsA = (endIndex + beginIndex_as_0) / 2 stepsA = (min(len(A), k) -1)/ 2 # stepsB = k - (stepsA + 1) -1 for the 0-based index stepsB = k - stepsA - 2 # Only array B contains elements if len(A) == 0: return B[k-1] # Both A and B contain elements, and we need the smallest one elif k == 1: return min(A[0], B[0]) # The median would be either A[stepsA] or B[stepsB], while A[stepsA] and # B[stepsB] have the same value. elif A[stepsA] == B[stepsB]: return A[stepsA] # The median must be in the right part of B or left part of A elif A[stepsA] > B[stepsB]: return findKthItem(A, B[stepsB+1:], k-stepsB-1) # The median must be in the right part of A or left part of B else: return findKthItem(A[stepsA+1:], B, k-stepsA-1) # There must be at least one element in these two arrays assert not(len(A) == 0 and len(B) == 0) if (len(A)+len(B))%2==1: # There are odd number of elements in total. The median the one in the middle return findKthItem(A, B, (len(A)+len(B))/2+1) * 1.0 else: # There are even number of elements in total. The median the mean value of the # middle two elements. return ( findKthItem(A, B, (len(A)+len(B))/2+1) + findKthItem(A, B, (len(A)+len(B))/2) ) / 2.0 |