4. Median of Two Sorted Arrays

'''
给出两个已排序数组（从小到大），求两个数组的中位数。

思路：
已排序的数组中位数应该很好求，通过索引计算就可以得到。两个数组的话就把它们合并起来再排序，然后很容易就得到了。不过题目有时间复杂度限制：O(log(m+n)).
这对排序算法就有了限制，不过用Python自带的排序函数，因为已经优化过了，所以运行时间不会超，就是有点投机取巧……
'''
# solution 1
class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        a = nums1 + nums2
        a.sort()
        length = len(a)
        if length%2 == 1:
            return a[(length-1)//2]*1.0
        else: return (a[length//2-1]+a[length//2])/2.0
        
# solution 2 （官方答案，用到中位数的作用，把数组分成两个部分，每个部分长度相等且max(left)<=min(right)）
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