1 Description #

source: https://leetcode.com/problems/search-in-rotated-sorted-array/

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, 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
• -10^4 <= nums[i] <= 10^4
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10^4 <= target <= 10^4

2 Solution #

class Solution {
public:
    int search(vector<int>& nums, int target) {
	int low = 0;
	int high = nums.size() - 1;
	while(low <= high){
	    int mid = (low + high) / 2;
	    if(nums[mid] == target){
		return mid;
	    }

	    // the left half is sorted
	    if(nums[mid] >= nums[low]){
		if(target <= nums[mid] && nums[low] <= target ){
		    high = mid - 1;
		}else{
		    low = mid + 1;
		}
	    }else{
		if(target <= nums[high] && nums[mid] <= target){
		    low = mid + 1;
		}else{
		    high = mid - 1;
		}
	    }
	}

	return -1;
    }
};