Search in Rotated Sorted Array

最新推荐文章于 2024-05-03 20:14:24 发布

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本文介绍了一种在已知没有重复元素的旋转数组中查找特定目标值的方法。通过使用改进后的二分查找法，该算法能在O(log n)的时间复杂度内找到目标值的位置或返回-1。文章详细阐述了如何利用旋转数组的特性来实现这一高效查找。

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Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

public class Solution {
    /**the roated list will be splitted into two parts after the rotate,and both are sorted list */
    /*we can use the lsearch, but here we can use the binary search with the attribute of rotated list */
    /*when we split the list with middle, there must be one sorted list */

    //assume no duplicate exists in the array
    public int search(int[] nums, int target) {
        if(nums == null || nums.length == 0)
            return -1;
        int low = 0;
        int high = nums.length - 1;
        while(low <= high) {
            int mid = (low + high);
            if(nums[mid] == target)
                return mid;

            if(nums[mid] > nums[low]) {
                if(target >= nums[low] && target < nums[mid])
                    high = mid - 1;
                else
                    low = mid + 1;
            }
            else {
                if(target > nums[mid] && target <= nums[high])
                    low = mid + 1;
                else
                    high = mid - 1;
            }
        }

        return -1;
    }
}