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
- All values of
nums
are unique. nums
is an ascending array that is possibly rotated.-104 <= target <= 104
Solution: Binary Search
If the current range [l, r] is ordered, reduce to normal binary search. Otherwise, determine the range to search next by comparing target and nums[0].
Time complexity: O(logn)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
// Author: Huahua class Solution { public: int search(vector<int>& nums, int target) { int l = 0; int r = nums.size(); while (l < r) { int m = l + (r - l) / 2; int x = (nums[m] < nums[0]) == (target < nums[0]) ? nums[m] : target < nums[0] ? INT_MIN : INT_MAX; if (x < target) l = m + 1; else if (x > target) r = m; else return m; } return -1; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment