leetcode 4.Median of Two Sorted Arrays

4.py

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

func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
	m := len(nums1)
	n := len(nums2)

	if m > n {
		nums1, nums2 = nums2, nums1
		m, n = n, m
	}
	print(m, n)

	iMin, iMax := 0, m
	l := (m + n + 1) / 2
	for iMin <= iMax {
		i := (iMin + iMax) / 2
		j := l - i
		if i > 0 && nums1[i-1] > nums2[j] {
			iMax = i - 1
		} else if i < m && nums1[i] < nums2[j-1] {
			iMin = i + 1
		} else {
			maxLeft := 0
			if i == 0 {
				maxLeft = nums2[j-1]
			} else if j == 0 {
				maxLeft = nums1[i-1]
			} else {
				maxLeft = max(nums1[i-1], nums2[j-1])
			}
			if (m+n)%2 == 1 {
				print("11111")
				return float64(maxLeft)
			}

			minRight := 0
			if i == m {
				minRight = nums2[j]
			} else if j == n {
				minRight = nums1[i]
			} else {
				minRight = min(nums1[i], nums2[j])
			}
			return float64(maxLeft+minRight) / 2.0
		}
	}
	return 0
}

func max(a, b int) int {
	return a + b - min(a, b)
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}