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33. Search in Rotated Sorted Array

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

Approach

Search in the rotated sorted array, we need to decide our search range, like binary search. We first define reverse point be the entry where

\[a[i - 1] > a[i] \text{ and } a[i + 1] > a[i]\]

  • If nums[low] <= nums[mid]

  • If nums[low] > nums[mid], the reverse point lie in the range.

Complexity

  • Time complexity: \(O(\log n)\)
  • Space complexity: \(O(1)\)

Code

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class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 1) {
            return nums[0] == target ? 0 : -1;
        }

        int low = 0; 
        int high = nums.length - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (nums[mid] == target) {
                return mid;
            }

            if (nums[low] <= nums[mid]) {
                if (nums[low] <= target && target < nums[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[high]) {
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
            // mid = (low + high) / 2;
        }
        return -1;
    }
}