leetcode

problem 4

4. Median of Two Sorted Arrays

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 $O(log (m+n))$.

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

答案要求复杂度为$O(log (m+n))$很容易可以想到我们需要用二分法来解决这个问题

其次问题是找中值，所以我们需要将数据切成两份

nums1: 1 2 3 4 5 6 7 8 9 (9)

: $\uparrow$

nums2: 2 3 5 8 9 (5)

: $\uparrow$

nums1: 1 2 3 4 5 6 7 8 9 (9)

: $\uparrow$

nums2: 2 3 5 8 9 (5)

: $\uparrow$

nums1: 1 2 3 4 5 6 7 8 9 (9)

: $\uparrow$

nums2: 2 3 5 8 9 (5)

: $\uparrow$

大致的问题解决了，下面需要思考的问题是奇偶数的问题，下标越界的问题

condition 1:

nums1: 1 2 3 4 5 6 7 8 9 (9)

: $\uparrow$

nums2: 2 3 5 8 9 (5)

condition 2:

nums1: 1 2 3 4 5 6 7 8 (8)

: $\uparrow$

nums2: 2 3 5 8 9 (5)

: $\uparrow$

class Solution:
    def findMedianSortedArrays(self,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, int((m + n + 1) / 2)
        while imin <= imax:
            i = int((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
        