LeetCode.33

最新推荐文章于 2023-05-21 20:40:47 发布

EdVulcan

最新推荐文章于 2023-05-21 20:40:47 发布

阅读量189

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode C/C++ 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_25371205/article/details/77966867

C/C++ 同时被 2 个专栏收录

19 篇文章

订阅专栏

LeetCode

16 篇文章

订阅专栏

本文介绍了一种在旋转过的有序数组中查找目标值的方法。通过递归地将数组分为两部分并确定目标值可能存在的区间来缩小搜索范围，最终实现高效查找。

Search in Rotated Sorted Array

Suppose an array sorted in ascending order 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.

int half(vector<int>& nums, int target, int begin, int end){
	int p = (begin + end) / 2;
	if(nums[begin] == target) return begin;
	if(nums[p] == target) return p;
	if(nums[end] == target) return end;
	if(nums[begin] < nums[p]){
		if(nums[begin] < target && target < nums[p]){
			return half(nums, target, begin, p);
		}
		else{
			return half(nums, target, p, end);
		}
	}
	else if(nums[begin] == nums[p]){
		return -1;
	}
	if(nums[p] < nums[end]){
		if(nums[p] < target && target < nums[end]){
			return half(nums, target, p, end);
		}
		else{
			return half(nums, target, begin, p);
		}
	}
	return -1;
}

int search(vector<int>& nums, int target){
	if(nums.empty()){
		return -1;
	}
	return half(nums, target, 0, nums.size() - 1);
}