Search in Rotated Sorted Array II

最新推荐文章于 2021-10-14 20:56:08 发布

liuyongvs2009

最新推荐文章于 2021-10-14 20:56:08 发布

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分类专栏： Leetcode

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本文讨论了含有重复元素的旋转有序数组搜索问题，并提供了一种有效的解决方案。通过自定义搜索算法，即使数组中存在重复元素也能找到目标值。文章详细解释了算法思路及其实现过程。

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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.

public class Solution {
    /*it is a good question following the rotated sorted array 1*/
    /*eg 3  1 2 3 3 3, we don;t know where to go,left or right. so we just start ++ until they are different.*/
    //if all the elmement is same, and the target is not in the array.the time complexity is O(n)
    public boolean search(int[] nums, int target) {
        if(nums == null || nums.length == 0)
            return false;
        int low = 0;
        int high = nums.length - 1;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(nums[mid] == target)
                return true;
            if(nums[mid] > nums[low]) {
                if(target >= nums[low] && target < nums[mid])
                    high = mid -1;
                else
                    low = mid + 1;
            }
            else if(nums[mid] < nums[low]) {
                if(target > nums[mid] && target <= nums[high])
                    low = mid + 1;
                else
                    high = mid -1;
            }
            //here is most important to solve the duplicate
            else  {
                low ++;
            }
        }

        return false;
    }
}