LeetCode | 33. Search in Rotated Sorted Array

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本文介绍了一种在旋转排序数组中查找目标值的算法。该数组原本按升序排列，然后在某个未知位置进行了旋转。通过递归实现二分查找，确保算法的时间复杂度为O(log n)，并提供两个示例说明其工作原理。

题外话：

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题目：

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

代码：

class Solution {
public:
    int backSearch(vector<int>& nums, int target, int low, int high){
		if(low > high)
			return -1;
		int mid = (low + high) / 2;
		if(nums[mid] == target)
			return mid;
		int mid_l = backSearch(nums, target, low, mid - 1);
		int mid_r = backSearch(nums, target, mid+1, high);
		
		if(mid_l == -1)
			return mid_r;
		return mid_l;
	}

	int search(vector<int>& nums, int target) {
        int low = 0, high = nums.size() - 1;
		return backSearch(nums, target, low, high);
    }
};