81. Search in Rotated Sorted Array II
- 本题难度: Medium
- Topic: Binary Search
类似[Lintcode]62. Search in Rotated Sorted Array/[Leetcode]33. Search in Rotated Sorted Array
Description
- Search in Rotated Sorted Array II
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
You are given a target value to search. If found in the array return true, otherwise return false.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:
This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
我的的代码
class Solution:
def search(self, A: \'List[int]\', target: \'int\') -> \'bool\':
# write your code here
if A == []:
return False
l = 0
r = len(A)-1#注意！
f = -1 # breakpoint
if len(A) == 1:
return A[0] == target
for i in range(1,len(A)):
if A[i-1]>A[i]:
f = i
if f == -1:
l = 0
r = len(A) - 1
else:
if target>=A[0]:
l = 0
r = f-1
else:
l = f
r = len(A)-1
while(l<=r):
m = (l+r)//2
if A[m] == target:
return True
else:
if A[m]>target:
r = m-1
else:
l = m+1
return False
思路
如果遇到[2,2,2,2,0,2,2]这类情况，没想好怎么做，所以
- 时间复杂度: O(log(n))
别人的代码
def search(self, nums, target):
l, r = 0, len(nums)-1
while l <= r:
mid = l + (r-l)//2
if nums[mid] == target:
return True
while l < mid and nums[l] == nums[mid]: # tricky part
l += 1
# the first half is ordered
if nums[l] <= nums[mid]:
# target is in the first half
if nums[l] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
# the second half is ordered
else:
# target is in the second half
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid - 1
return False
思路
如果遇到[2,2,2,2,0,2,2]这类情况，没想好怎么做，所以
- 时间复杂度: average : O(log(n)) worst O(n)