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There is an integer array nums
sorted in ascending order (with distinct values).
Prior to being passed to your function, nums
is possibly rotated at an unknown pivot 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 rotated at pivot index 3
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
nums
are unique.nums
is an ascending array that is possibly rotated.-104 <= target <= 104
public class Solution {
public int search(int[] A, int target) {
int lo = 0;
int hi = A.length - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (A[mid] == target) return mid;
if (A[lo] <= A[mid]) {
if (target >= A[lo] && target < A[mid]) {
hi = mid - 1;
} else {
lo = mid + 1;
}
} else {
if (target > A[mid] && target <= A[hi]) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
}
return A[lo] == target ? lo : -1;
}
class Solution:
# @param {integer[]} numss
# @param {integer} target
# @return {integer}
def search(self, nums, target):
if not nums:
return -1
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) / 2
if target == nums[mid]:
return mid
if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1
return -1
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (l+r) / 2;
if (target == nums[mid])
return mid;
// there exists rotation; the middle element is in the left part of the array
if (nums[mid] > nums[r]) {
if (target < nums[mid] && target >= nums[l])
r = mid - 1;
else
l = mid + 1;
}
// there exists rotation; the middle element is in the right part of the array
else if (nums[mid] < nums[l]) {
if (target > nums[mid] && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}
// there is no rotation; just like normal binary search
else {
if (target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}
}
return -1;
}
};
def search(nums, target)
i = (0...nums.size).bsearch { |i|
(nums[0] <= target) ^ (nums[0] > nums[i]) ^ (target > nums[i])
}
nums[i || 0] == target ? i : -1
end
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