81. Search in Rotated Sorted Array II

最新推荐文章于 2023-11-12 22:36:10 发布

SnailRunFast

最新推荐文章于 2023-11-12 22:36:10 发布

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本文探讨了在一个未知旋转点的升序数组中搜索目标值的问题，提供了详细的算法实现，包括处理数组可能包含重复元素的情况，讨论了该算法的时间复杂度。

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81. Search in Rotated Sorted Array II

Medium

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:

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

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?

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