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There are two sorted arrays nums1 and nums2 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)).
class Solution(object): def findMedianSortedArrays(self, nums1, nums2): new_list = [] median = 0.0 l = 0 new_list = nums1 + nums2 l = len(new_list) new_list.sort() if (l%2) == 0: l = int(l/2) median = float((new_list[l-1]+new_list[l])/2) else: l = l//2 median = new_list[l] return median
But this solution's O(n^2)
A simpler on of O(m+n):
class Solution(object): def findMedianSortedArrays(self, nums1, nums2): new_list = [] median = 0.0 l = 0 while min(len(nums1),len(nums2)): if nums1[0]<nums2[0]: new_list.append(nums1[0]) del nums1[0] if len(nums1)==0: break if nums1[0]>nums2[0]: new_list.append(nums2[0]) del nums2[0] if len(nums2)==0: break else: new_list.append(nums1[0]) new_list.append(nums2[0]) del nums1[0] del nums2[0] if len(nums1)==0 or len(nums2)==0: break if len(nums1): for i in range(len(nums1)): new_list.append(nums1[i]) if len(nums2): for i in range(len(nums2)): new_list.append(nums2[i]) l = len(new_list) if (l%2) == 0: l = int(l/2) median = float((new_list[l-1]+new_list[l])/2) else: l = l//2 median = new_list[l] return median
Debug experiece:
1--
Version 0.0:
while len(nums1) & len(nums2): if nums1[0]<nums2[0]: new_list.append(nums1[0]) del nums1[0] if len(nums1)==0: break ............
Please becareful about using & or 'and' in python:
>>> 1&2 0 >>> 2&4 0 >>> 3&4 0 >>> 1&1 1 >>> 2 and 1 1 >>> 2 and 3 3
You can find the calculation is processing with binary code.
Version 0.1 :
while min(len(nums1),len(nums2)): if nums1[0]<nums2[0]: new_list.append(nums1[0]) del nums1[0] ## if len(nums1)==0: ## break if nums1[0]>nums2[0]: new_list.append(nums2[0]) del nums2[0] ## if len(nums2)==0: ## break else: new_list.append(nums1[0]) new_list.append(nums2[0]) del nums1[0] del nums2[0] ## if len(nums1)==0 or len(nums2)==0: ## break
When you try to running with test cases: [], [2,3,4]
it will show the error mentioning about out of index__but it's fine to run with python shell. Seems in leetcode, it will run the codes without checking the criterial.
Solution_Recursive Approach
def median(A, B): m, n = len(A), len(B) if m > n: A, B, m, n = B, A, n, m if n == 0: raise ValueError imin, imax, half_len = 0, m, (m + n + 1) / 2 while imin <= imax: i = (imin + imax) / 2 j = half_len - i if i < m and B[j-1] > A[i]: # i is too small, must increase it imin = i + 1 elif i > 0 and A[i-1] > B[j]: # i is too big, must decrease it imax = i - 1 else: # i is perfect if i == 0: max_of_left = B[j-1] elif j == 0: max_of_left = A[i-1] else: max_of_left = max(A[i-1], B[j-1]) if (m + n) % 2 == 1: return max_of_left if i == m: min_of_right = B[j] elif j == n: min_of_right = A[i] else: min_of_right = min(A[i], B[j]) return (max_of_left + min_of_right) / 2.0
More details :
https://discuss.leetcode.com/topic/4996/share-my-o-log-min-m-n-solution-with-explanation/2
Analysis in Chinese:
http://blog.csdn.net/hk2291976/article/details/51107778