Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Your algorithm‘s runtime complexity must be in the order of O(log n).
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: -133. Search in Rotated Sorted Array
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 (target > nums[mid] && target <= nums[r]) l = mid + 1; else r = mid - 1; } } return -1; } };
33.[LeetCode] Search in Rotated Sorted Array
原文：https://www.cnblogs.com/250101249-sxy/p/10440472.html