LeetCode 081 Search in Rotated Sorted Array II

最新推荐文章于 2020-06-27 10:28:57 发布

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本文针对Search in Rotated Sorted Array的跟进问题，探讨了当数组中允许存在重复元素时，查找指定目标值的解决方案。文章提供了一种有效的算法实现，并分析了这种情况下算法的时间复杂度变化。

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题目描述

Follow up for “Search in Rotated Sorted Array”:
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

代码

    public boolean search(int[] nums, int target) {

        int left = 0, right = nums.length - 1;

        while (left <= right) {

            int mid = (left + right) / 2;

            if (nums[mid] == target)
                return true;

            if (nums[left] < nums[mid]) {
                if (target <= nums[mid] && target >= nums[left])
                    right = mid - 1;
                else
                    left = mid + 1;
            } else if (nums[left] > nums[mid]) {
                if (target >= nums[left] || target <= nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            } else
                left++;
        }

        return false;
    }