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 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

Train of thought

• The time complexity is requirement O(log n), so finding the median of the array is simple way to solve it.
• Using compared way with the median of the array to decide which part should be required in the next compared round.
• Whilst move the low and high pointer to find, the low pointer is started by index 0, and the high pointer is started by the last index in the array

Solution

public static int search(int[] nums, int target) {
    int low = 0;
    int high = nums.length - 1;
    while (low <= high) {
        // find the median element of the array
        int mid = (low + high) / 2;
        if (nums[mid] == target) {
            return mid;
        }
        if (nums[low] <= nums[mid]) {
            if (target >= nums[low] && target <= nums[mid]) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        } else {
            if (target >= nums[mid] && target <= nums[high]) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
    }
    return -1;
}

The complexity

Time complexity: O(log n)