LeetCode 33：搜索旋转排序数组（Search in Rotated Sorted Array）解法汇总

最新推荐文章于 2021-02-20 07:29:54 发布

Ubuntu1996

最新推荐文章于 2021-02-20 07:29:54 发布

阅读量226

点赞数

分类专栏： LeetCode刷题题解记录文章标签： LeetCode 搜索旋转排序数组 Search in Rotated Sorted Array

本文链接：https://blog.youkuaiyun.com/Utopiabuntu/article/details/97138420

版权

LeetCode刷题题解记录专栏收录该内容

54 篇文章

订阅专栏

这篇博客探讨了LeetCode 33题——搜索旋转排序数组的问题，要求算法复杂度为O(log n)。文章通过引入二分查找策略，并针对旋转排序数组的特点增加判断条件，提出了解决方案。并给出了两个示例输入和相应的输出结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

文章目录

Solution

更多LeetCode题解
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.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Solution

看到复杂度 $O (l o g (n))$ ，就想到了二分查找法，此题是二分查找的一个变体，多加一些判断条件就好了。

class Solution {
public:
	int search(vector<int>& nums, int target) {
		int i = 0, j = nums.size()-1;
		if (j == -1) return -1;
		while (i < j) {
			int m = (i + j) / 2;
			if (target >= nums[i]) {
				if (target <= nums[m]) {
					j = m;
				}
				else {
					if (nums[m] >= nums[i]) {
						i = m + 1;
					}
					else {
						j = m;
					}
				}
			}
			else if (target <= nums[j]) {
				if (target >= nums[m]) {
					i = m;
				}
				else {
					if (nums[m] < nums[j]) {
						j = m;
					}
					else {
						i = m + 1;
					}
				}
			}
			else {
				return -1;
			}
		}
		if (nums[i] == target)
			return i;
		else
			return -1;
	}
};