leetcode-4.py

# -*- coding:utf-8 -*-

"""
    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
"""

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


"""
    解决关键:
    分割：
    
             left_A             |        right_A
    A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]
    
    
              left_B             |        right_B
    B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]
    
    合并：
    
              left_part          |        right_part
    A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]
    B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]
    
    
    关键：
    len(left_part)=len(right_part)
    max(left_part)≤min(right_part)
"""