标签：ret pre while return tps sort float == else
00
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
Solution with Algorithms mind:
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://leetcode.com/articles/median-of-two-sorted-arrays/
leetcode -- Algorithms -- 4_ Median of Two Sorted Arrays
标签：ret pre while return tps sort float == else
原文地址：http://www.cnblogs.com/Shareishappy/p/7506935.html
