Leetcode no.33

最新推荐文章于 2019-04-03 16:12:23 发布

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本文介绍了一种在旋转过的有序数组中查找特定元素的方法。通过实现一个高效的搜索算法，该算法能在O(log n)的时间复杂度内找到目标值的索引位置，如果未找到则返回-1。文章提供了一个具体的Java代码示例。

33. Search in Rotated Sorted Array

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 {
    public int search(int[] nums, int target) {
        if (nums.length==0) return -1;
        int start= 0, end= nums.length-1;
        while (start <= end){
            int center= (start+end)/2;
            if (nums[center] == target) return center;
            if (nums[center]>= nums[start]){
                if (target < nums[center] && target>= nums[start])
                    end= center-1;
                else start= center+1;
            }
            if (nums[center]<=nums[end]){
                if (target > nums[center] && target<=nums[end])
                    start= center+1;
                else end= center-1;
            }
        }
        return -1;
    }
}