(Medium #33) Search in Rotated Sorted Array
Problem
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7]
might become [4,5,6,7,0,1,2]
).
You are given a target value to search. If found in the array return its index, otherwise return -1
.
You may assume no duplicate exists in the array.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2]
, target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2]
, target = 3
Output: -1
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e.,
[0,1,2,4,5,6,7]
might become [4,5,6,7,0,1,2]
).
You are given a target value to search. If found in the array return its index, otherwise return
-1
.
You may assume no duplicate exists in the array.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2]
, target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2]
, target = 3
Output: -1
My solution
# https://leetcode.com/problems/search-in-rotated-sorted-array/discuss/145697/Python-100
class Solution:
def search(self, nums, target):
s, e = 0, len(nums)-1
while s <= e:
m = int((s+e) / 2)
if nums[m] == target:
return m
elif nums[m] > target:
if nums[m] >= nums[s]:
if nums[s] > target:
s = m+1
else: ## nums[s] < target
e = m-1
else: ## nums[m] < nums[s]
e = m-1
else: ## nums[m] < target
if nums[m] >= nums[s]:
s = m+1
else: ## nums[m] < nums[s]
if nums[e] >= target:
s = m+1
else: ## nums[e] < target
e = m-1
return -1
Runtime: 52ms, Memory: 14MB
- 참고(index함수이용): Runtime: 48ms, Memory: 14.3MB
- why not using index() ? is it allowed?
- I think this is because it will be O(n) solution, not O(logn) as in condition
class Solution:
def search(self, nums, target):
s, e = 0, len(nums)-1
while s <= e:
m = int((s+e) / 2)
if nums[m] == target:
return m
elif nums[m] > target:
if nums[m] >= nums[s]:
if nums[s] > target:
s = m+1
else: ## nums[s] < target
e = m-1
else: ## nums[m] < nums[s]
e = m-1
else: ## nums[m] < target
if nums[m] >= nums[s]:
s = m+1
else: ## nums[m] < nums[s]
if nums[e] >= target:
s = m+1
else: ## nums[e] < target
e = m-1
return -1
- 참고(index함수이용): Runtime: 48ms, Memory: 14.3MB
- why not using index() ? is it allowed?
- I think this is because it will be O(n) solution, not O(logn) as in condition
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