LeetCode题解：0004
题目链接（困难）
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 | 内存消耗 |
|---|---|---|---|---|
| Ans 1 (Python) | O ( ( M + N ) l o g ( M + N ) ) O((M+N)log(M+N)) O((M+N)log(M+N)) | O ( M + N ) O(M+N) O(M+N) | 40ms (>98.73%) | 13.7MB (>6.15%) |
| Ans 2 (Python) | O ( M + N ) O(M+N) O(M+N) | O ( 1 ) O(1) O(1) | 48ms (>89.95%) | 13.7MB (>6.15%) |
| Ans 3 (Python) | O ( l o g ( M + N ) ) O(log(M+N)) O(log(M+N)) | O ( 1 ) O(1) O(1) | 80ms (>11.71%) | 13.8MB (>6.15%) |
解法一（直接合并数组排序求中位数）：
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
nums = nums1 + nums2
nums.sort()
count = len(nums)
if count % 2 == 0:
return (nums[int(count / 2 - 1)] + nums[int(count / 2)]) / 2
else:
return nums[int(count / 2)]
解法二（直接寻找中位数的位置）：
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
len_1 = len(nums1)
len_2 = len(nums2)
total = len_1 + len_2
left = -1
right = -1
index_1 = 0
index_2 = 0
for i in range(int(total / 2) + 1):
left = right
if index_1 < len_1 and (index_2 >= len_2 or nums1[index_1] < nums2[index_2]):
right = nums1[index_1]
index_1 += 1
else:
right = nums2[index_2]
index_2 += 1
if total % 2 == 0:
return (left + right) / 2
else:
return right
解法三（使用二分查找寻找中位数的位置）：
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
len_1 = len(nums1)
len_2 = len(nums2)
total = len_1 + len_2
if len_1 == 0:
if total % 2 == 0:
return (nums2[int(total / 2 - 1)] + nums2[int(total / 2)]) / 2
else:
return nums2[int(total / 2)]
if len_2 == 0:
if total % 2 == 0:
return (nums1[int(total / 2 - 1)] + nums1[int(total / 2)]) / 2
else:
return nums1[int(total / 2)]
middle_num = (total + 1) // 2
index_1 = -1
index_2 = -1
already_num = 0
while True:
surplus_num = middle_num - already_num # 剩余总量
if surplus_num == 1:
find_number = 1 # 这一次二分的数量
else:
find_number = surplus_num // 2 # 这一次二分的数量
binery_index_1 = index_1 + find_number
binery_index_2 = index_2 + find_number
# print(binery_index_1, binery_index_2)
if binery_index_1 >= len_1:
binery_index_1 = len_1 - 1
if binery_index_2 >= len_2:
binery_index_2 = len_2 - 1
if index_1 == len_1 - 1:
already_num += binery_index_2 - index_2
index_2 = binery_index_2
if index_2 == len_2 - 1:
already_num += binery_index_1 - index_1
index_1 = binery_index_1
# print(binery_index_1, binery_index_2)
if nums1[binery_index_1] < nums2[binery_index_2]:
already_num += binery_index_1 - index_1
index_1 = binery_index_1
else:
already_num += binery_index_2 - index_2
index_2 = binery_index_2
print("List1[" + str(binery_index_1) + "] =", nums1[binery_index_1], ",",
"List2[" + str(binery_index_2) + "] =", nums2[binery_index_2],
"→", min(nums1[binery_index_1], nums2[binery_index_2]), "(", already_num, ")", "-",
index_1, index_2)
if already_num >= middle_num:
break
print("index_1 =", index_1, "; index_2 =", index_2)
if index_1 == -1:
now_number = nums2[index_2]
# print("Now:", index_2, nums2[index_2])
elif index_2 == -1:
now_number = nums1[index_1]
# print("Now:", index_1, nums1[index_1])
else:
now_number = max(nums1[index_1], nums2[index_2])
# print("Now:", index_1, nums1[index_1], ",", index_2, nums2[index_2])
if index_1 + 1 >= len_1:
next_number = nums2[index_2 + 1]
elif index_2 + 1 >= len_2:
next_number = nums1[index_1 + 1]
else:
next_number = min(nums1[index_1 + 1], nums2[index_2 + 1])
print("Now:", now_number, ";Next:", next_number)
if total % 2 == 1:
return now_number
else:
return (now_number + next_number) / 2
