Solve using dichotomy
- Step1, k = (n + m) / 2
After merging the two arrays, take the median
– if n + m is even, take the smallest K, K + 1
– if n + m is odd, take the first K small number - Step2, compare the K /
2th numbers of the two arrays and round down. If the number in num1 is smaller than the number in num2, then remove the first K / 2 numbers in num1 - Step3: Update the K value. In the new num1 and num2 arrays, continue to remove the first K / 2 numbers.
Use the dichotomy method to solve.
If one of the arrays is empty, then select the rest from another array.
If K = 1, Then from the two arrays, find the smallest one
nums1 = [1, 3, 5, 7]
nums2 = [1,2,3,4,5,6,7,8,9,10]
because n + m is even, median = (7th smallest + 8th smallest) / 2
"""
寻找两个有序数组的中位数
利用二分查找的优势，求解
"""
# 在nums1[start1,end1], nums2[start2, end2]两个数组中，找第k小的数
def get_kth_num(nums1, start1, end1, nums2, start2, end2, k):
len1 = end1 - start1 + 1 # 第一个数组的长度
len2 = end2 - start2 + 1 # 第二个数组的长度
# 让 len1 的长度小于 len2，这样就能保证如果有数组空了，一定是 len1
if len1 > len2: # 这里想要保住len1的长度比len2的长度小，如果len1长，则把len1和len2颠倒下
return get_kth_num(nums2, start2, end2, nums1, start1, end1, k)
if len1 == 0: # 此时len2（那个数比较多的数组）还是有值，但是len1已经空了
return nums2[start2 + k - 1]
if k == 1:
return min(nums1[start1], nums2[start2]) # 用一个min函数找两者中较小的那个
# 两个数组，分别前进k/2个数
i = int(start1 + min(len1, k / 2) - 1)
j = int(start2 + min(len2, k / 2) - 1)
if nums1[i] < nums2[j]:
# 第一个数组，前k/2去掉，在剩余的数组中，继续查找
return get_kth_num(nums1, i + 1, end1, nums2, start2, end2, k - (i - start1 + 1))
else:
# 第二个数组，前k/2去掉，在剩余的数组中，继续查找
return get_kth_num(nums1, start1, end1, nums2, j + 1, end2, k - (j - start2 + 1))
# 外层函数：找两个有序数组中的中位数
def find_median(nums1, nums2):
n = len(nums1)
m = len(nums2)
# 第一个中位数
left = int((n + m + 1) / 2)
# 第二个中位数
right = int((n + m + 2) / 2)
# 以下分支结构实现两个数组奇数个找出来，偶数个找出两个数再取平均数
if left == right:
result = get_kth_num(nums1, 0, n - 1, nums2, 0, m - 1, left)
else:
result = (get_kth_num(nums1, 0, n - 1, nums2, 0, m - 1, left) + get_kth_num(nums1, 0, n - 1, nums2, 0, m - 1, right)) / 2
return result
"""
nums1 = [1, 2, 3]
nums2 = [4, 5]
"""
nums1 = [1, 3, 5, 7]
nums2 = [1,2,3,4,5,6,7,8,9,10]
print (find_median(nums1, nums2))
