Description
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly left rotated at an unknown index k (1 <= 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, [0,1,2,4,5,6,7] might be left rotated by 3 indices and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
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
Example 3:
Input: nums = [1], target = 0
Output: -1
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104- All values of
nums are unique.
nums is an ascending array that is possibly rotated.-104 <= target <= 104
Solutions
Solution 1: Binary Search
We use binary search to divide the array into two parts, [left,.. mid] and [mid + 1,.. right]. At this point, we can find that one part must be sorted.
Therefore, we can determine whether target is in this part based on the sorted part:
- If the elements in the range [0,.. mid] form a sorted array:
- If nums[0] ≤ target ≤ nums[mid], then our search range can be narrowed down to [left,.. mid];
- Otherwise, search in [mid + 1,.. right];
- If the elements in the range [mid + 1, n - 1] form a sorted array:
- If nums[mid] \lt target ≤ nums[n - 1], then our search range can be narrowed down to [mid + 1,.. right];
- Otherwise, search in [left,.. mid].
The termination condition for binary search is left ≥ right. If at the end we find that nums[left] is not equal to target, it means that there is no element with a value of target in the array, and we return -1. Otherwise, we return the index left.
The time complexity is O(log n), where n is the length of the array nums. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScriptC#PHP
class Solution:
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
left, right = 0, n - 1
while left < right:
mid = (left + right) >> 1
if nums[0] <= nums[mid]:
if nums[0] <= target <= nums[mid]:
right = mid
else:
left = mid + 1
else:
if nums[mid] < target <= nums[n - 1]:
left = mid + 1
else:
right = mid
return left if nums[left] == target else -1(code-box)
class Solution {
public int search(int[] nums, int target) {
int n = nums.length;
int left = 0, right = n - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[0] <= nums[mid]) {
if (nums[0] <= target && target <= nums[mid]) {
right = mid;
} else {
left = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[n - 1]) {
left = mid + 1;
} else {
right = mid;
}
}
}
return nums[left] == target ? left : -1;
}
}(code-box)
class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
int left = 0, right = n - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[0] <= nums[mid]) {
if (nums[0] <= target && target <= nums[mid])
right = mid;
else
left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[n - 1])
left = mid + 1;
else
right = mid;
}
}
return nums[left] == target ? left : -1;
}
};(code-box)
func search(nums []int, target int) int {
n := len(nums)
left, right := 0, n-1
for left < right {
mid := (left + right) >> 1
if nums[0] <= nums[mid] {
if nums[0] <= target && target <= nums[mid] {
right = mid
} else {
left = mid + 1
}
} else {
if nums[mid] < target && target <= nums[n-1] {
left = mid + 1
} else {
right = mid
}
}
}
if nums[left] == target {
return left
}
return -1
}(code-box)
function search(nums: number[], target: number): number {
const n = nums.length;
let left = 0,
right = n - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[0] <= nums[mid]) {
if (nums[0] <= target && target <= nums[mid]) {
right = mid;
} else {
left = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[n - 1]) {
left = mid + 1;
} else {
right = mid;
}
}
}
return nums[left] == target ? left : -1;
}(code-box)
impl Solution {
pub fn search(nums: Vec<i32>, target: i32) -> i32 {
let mut l = 0;
let mut r = nums.len() - 1;
while l <= r {
let mid = (l + r) >> 1;
if nums[mid] == target {
return mid as i32;
}
if nums[l] <= nums[mid] {
if target < nums[mid] && target >= nums[l] {
r = mid - 1;
} else {
l = mid + 1;
}
} else {
if target > nums[mid] && target <= nums[r] {
l = mid + 1;
} else {
r = mid - 1;
}
}
}
-1
}
}(code-box)
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function (nums, target) {
const n = nums.length;
let left = 0,
right = n - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[0] <= nums[mid]) {
if (nums[0] <= target && target <= nums[mid]) {
right = mid;
} else {
left = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[n - 1]) {
left = mid + 1;
} else {
right = mid;
}
}
}
return nums[left] == target ? left : -1;
};(code-box)
public class Solution {
public int Search(int[] nums, int target) {
int n = nums.Length;
int left = 0, right = n - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[0] <= nums[mid]) {
if (nums[0] <= target && target <= nums[mid]) {
right = mid;
} else {
left = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[n - 1]) {
left = mid + 1;
} else {
right = mid;
}
}
}
return nums[left] == target ? left : -1;
}
}(code-box)
class Solution {
/**
* @param integer[] $nums
* @param integer $target
* @return integer
*/
function search($nums, $target) {
$foundKey = -1;
foreach ($nums as $key => $value) {
if ($value === $target) {
$foundKey = $key;
}
}
return $foundKey;
}
}(code-box)
