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] <= 104numsis 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 of the binary search as $l = 0$ and the right boundary as $r = n - 1$, where $n$ is the length of the array.
Each time during the binary search, we get the current midpoint $\textit{mid} = (l + r) / 2$.
If $\textit{nums}[\textit{mid}] > \textit{nums}[r]$, it means $[l, \textit{mid}]$ is ordered. If $\textit{nums}[l] \le \textit{target} \le \textit{nums}[\textit{mid}]$, it means $\textit{target}$ is in $[l, \textit{mid}]$. Otherwise, $\textit{target}$ is in $[\textit{mid} + 1, r]$.
If $\textit{nums}[\textit{mid}] < \textit{nums}[r]$, it means $[\textit{mid} + 1, r]$ is ordered. If $\textit{nums}[\textit{mid}] < \textit{target} \le \textit{nums}[r]$, it means $\textit{target}$ is in $[\textit{mid} + 1, r]$. Otherwise, $\textit{target}$ is in $[l, \textit{mid}]$.
If $\textit{nums}[\textit{mid}] = \textit{nums}[r]$, it means the elements $\textit{nums}[\textit{mid}]$ and $\textit{nums}[r]$ are equal. In this case, we cannot determine which interval $\textit{target}$ is in, so we can only decrease $r$ by $1$.
After the binary search, if $\textit{nums}[l] = \textit{target}$, it means the target value $\textit{target}$ exists in the array. Otherwise, it does not exist.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
Python3
class Solution:
def search(self, nums: List[int], target: int) -> bool:
n = len(nums)
l, r = 0, n - 1
while l < r:
mid = (l + r) >> 1
if nums[mid] > nums[r]:
if nums[l] <= target <= nums[mid]:
r = mid
else:
l = mid + 1
elif nums[mid] < nums[r]:
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid
else:
r -= 1
return nums[l] == target
Java
class Solution {
public boolean search(int[] nums, int target) {
int l = 0, r = nums.length - 1;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] > nums[r]) {
if (nums[l] <= target && target <= nums[mid]) {
r = mid;
} else {
l = mid + 1;
}
} else if (nums[mid] < nums[r]) {
if (nums[mid] < target && target <= nums[r]) {
l = mid + 1;
} else {
r = mid;
}
} else {
--r;
}
}
return nums[l] == target;
}
}
C++
class Solution {
public:
bool search(vector<int>& nums, int target) {
int l = 0, r = nums.size() - 1;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] > nums[r]) {
if (nums[l] <= target && target <= nums[mid]) {
r = mid;
} else {
l = mid + 1;
}
} else if (nums[mid] < nums[r]) {
if (nums[mid] < target && target <= nums[r]) {
l = mid + 1;
} else {
r = mid;
}
} else {
--r;
}
}
return nums[l] == target;
}
};
Go
func search(nums []int, target int) bool {
l, r := 0, len(nums)-1
for l < r {
mid := (l + r) >> 1
if nums[mid] > nums[r] {
if nums[l] <= target && target <= nums[mid] {
r = mid
} else {
l = mid + 1
}
} else if nums[mid] < nums[r] {
if nums[mid] < target && target <= nums[r] {
l = mid + 1
} else {
r = mid
}
} else {
r--
}
}
return nums[l] == target
}
TypeScript
function search(nums: number[], target: number): boolean {
let [l, r] = [0, nums.length - 1];
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] > nums[r]) {
if (nums[l] <= target && target <= nums[mid]) {
r = mid;
} else {
l = mid + 1;
}
} else if (nums[mid] < nums[r]) {
if (nums[mid] < target && target <= nums[r]) {
l = mid + 1;
} else {
r = mid;
}
} else {
--r;
}
}
return nums[l] === target;
}
Rust
impl Solution {
pub fn search(nums: Vec<i32>, target: i32) -> bool {
let (mut l, mut r) = (0, nums.len() - 1);
while l < r {
let mid = (l + r) >> 1;
if nums[mid] > nums[r] {
if nums[l] <= target && target <= nums[mid] {
r = mid;
} else {
l = mid + 1;
}
} else if nums[mid] < nums[r] {
if nums[mid] < target && target <= nums[r] {
l = mid + 1;
} else {
r = mid;
}
} else {
r -= 1;
}
}
nums[l] == target
}
}
JavaScript
/**
* @param {number[]} nums
* @param {number} target
* @return {boolean}
*/
var search = function (nums, target) {
let [l, r] = [0, nums.length - 1];
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] > nums[r]) {
if (nums[l] <= target && target <= nums[mid]) {
r = mid;
} else {
l = mid + 1;
}
} else if (nums[mid] < nums[r]) {
if (nums[mid] < target && target <= nums[r]) {
l = mid + 1;
} else {
r = mid;
}
} else {
--r;
}
}
return nums[l] === target;
};