81. Search in Rotated Sorted Array II
Description
There is an integer array nums
sorted in non-decreasing order (not necessarily with distinct values).
Before being passed to your function, nums
is rotated at an unknown pivot index k
(0 <= k < nums.length
) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]
(0-indexed). For example, [0,1,2,4,4,4,5,6,6,7]
might be rotated at pivot index 5
and become [4,5,6,6,7,0,1,2,4,4]
.
Given the array nums
after the rotation and an integer target
, return true
if target
is in nums
, or false
if it is not in nums
.
You must decrease the overall operation steps as much as possible.
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
Constraints:
1 <= nums.length <= 5000
-104 <= nums[i] <= 104
nums
is guaranteed to be rotated at some pivot.-104 <= target <= 104
Follow up: This problem is similar to Search in Rotated Sorted Array, but nums
may contain duplicates. Would this affect the runtime complexity? How and why?
Solutions
Solution 1: Binary Search
We define the left boundary $l=0$ and the right boundary $r=n-1$ for the binary search, where $n$ is the length of the array.
During each binary search process, we get the current midpoint $mid=(l+r)/2$.
- If $nums[mid] \gt nums[r]$, it means that $[l,mid]$ is ordered. At this time, if $nums[l] \le target \le nums[mid]$, it means that $target$ is in $[l,mid]$, otherwise $target$ is in $[mid+1,r]$.
- If $nums[mid] \lt nums[r]$, it means that $[mid+1,r]$ is ordered. At this time, if $nums[mid] \lt target \le nums[r]$, it means that $target$ is in $[mid+1,r]$, otherwise $target$ is in $[l,mid]$.
- If $nums[mid] = nums[r]$, it means that the elements $nums[mid]$ and $nums[r]$ are equal. At this time, we cannot determine which interval $target$ is in, so we can only decrease $r$ by $1$.
After the binary search ends, if $nums[l] = target$, it means that the target value $target$ exists in the array, otherwise it means it does not exist.
The time complexity is approximately $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array.
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