[LeetCode] 81. Search in Rotated Sorted Array II

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Problem

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

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

Example 2:

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

Follow up:

  • This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
  • Would this affect the run-time complexity? How and why?

Explanation

  1. Similar to 33. Search in Rotated Sorted Array, now, we also need to consider the case that nums[mid] == nums[left] == num[right]. In this case, we can move left one step forward, so the mid will also move forward, until !(nums[mid] == nums[left] == num[right]), and now it is the same as 33. Search in Rotated Sorted Array.

Solution

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class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) return false;
        int left = 0;
        int right = nums.length-1;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return true;
            } else if (nums[mid] < nums[left]) {
                if (target > nums[mid] && target <= nums[right]) {
                    left = mid+1;
                } else {
                    right = mid;
                }
            } else if (nums[mid] > nums[left]) {
                if (target >= nums[left] && target < nums[mid]) {
                    right = mid;
                } else {
                    left = mid+1;
                }
            } else {
                left += 1;
            }
        }

        if (nums[left] == target) {
            return true;
        } else {
            return false;
        }
    }
}