LeetCode 81. Search in Rotated Sorted Array II

Question

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

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

Solution

The differences between this problem and LeetCode 33. Search in Rotated Sorted Array are: 1. in this problem, nums may contain duplicate elements; 2. the solution only needs to return whether nums contains target, but not the index of target if it exists.

// Time Complexity: average O(log(n)); worst O(n)
public boolean search(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
        return false;
    }

    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return true;
        }
        if (nums[mid] > nums[left]) {
            // [left, mid] is sorted
            if (nums[mid] > target && nums[left] <= target) {
                // target may be in the sorted interval
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else if (nums[mid] < nums[left]) {
            // [mid, right] is sorted
            if (nums[mid] < target && nums[right] >= target) {
                // target may be in the sorted interval
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        } else {
            // nums[mid] == nums[left]
            // exclude left to make progress.
            // it is safe because nums[left] != target
            left++;
        }
    }
    return false;
}