33.[LeetCode] Search in Rotated Sorted Array

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;
    }
};