给定两个大小为 m 和 n 的有序数组 nums1
和 nums2
。
请你找出这两个有序数组的中位数,并且要求算法的时间复杂度为 O(log(m + n))。
你可以假设 nums1
和 nums2
不会同时为空。
示例 1:
nums1 = [1, 3]
nums2 = [2] 则中位数是 2.0
示例 2:
nums1 = [1, 2]
nums2 = [3, 4] 则中位数是 (2 + 3)/2 = 2.5 代码如下:
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
已测试通过