72 lines
1.3 KiB
Markdown
72 lines
1.3 KiB
Markdown
|
---
|
||
|
|
layout: page
|
||
|
|
title: Search in Rotated Sorted Array
|
||
|
|
last_solved: 2020-06-07
|
||
|
|
category: binary search
|
||
|
|
leetcode_url: https://leetcode.com/problems/search-in-rotated-sorted-array
|
||
|
|
status: Attempted
|
||
|
|
---
|
||
|
|
|
||
|
|
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
|
||
|
|
|
||
|
|
|
||
|
|
```
|
||
|
|
|
||
|
|
Solution
|
||
|
|
----------
|
||
|
|
|
||
|
|
Check Nick White's video on this question. it's also 2-pass binary search
|
||
|
|
|
||
|
|
{% highlight python %}
|
||
|
|
|
||
|
|
def search(self, nums, target):
|
||
|
|
if not nums:
|
||
|
|
return -1
|
||
|
|
l,r=0,len(nums)-1
|
||
|
|
while l<=r:
|
||
|
|
m=(l+r)/2
|
||
|
|
if nums[m]<nums[0]:
|
||
|
|
r=m-1
|
||
|
|
else:
|
||
|
|
l=m+1
|
||
|
|
|
||
|
|
if nums[0]>target:
|
||
|
|
r=len(nums)-1
|
||
|
|
else:
|
||
|
|
l=0
|
||
|
|
|
||
|
|
while l<=r:
|
||
|
|
m=(l+r)/2
|
||
|
|
if nums[m]==target:
|
||
|
|
return m
|
||
|
|
if nums[m]<target:
|
||
|
|
l=m+1
|
||
|
|
else:
|
||
|
|
r=m-1
|
||
|
|
return -1
|
||
|
|
|
||
|
|
{% endhighlight %}
|
||
|
|
|
||
|
|
|
||
|
|
![image1]()
|