LeetCode33:搜索旋转排序数组(Search in Rotated Sorted Array)

英文题目：

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

中文题目：

假设按照升序排序的数组在预先未知的某个点上进行了旋转。

( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。

搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。

你可以假设数组中不存在重复的元素。

你的算法时间复杂度必须是 O(log n) 级别。

示例 1:

输入: nums = [4,5,6,7,0,1,2], target = 0
输出: 4

示例 2:

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

解答：

C++

class Solution {
public:
    int search(vector& nums, int target) {
        if (nums.empty()) return -1;
        
        //找到最小值
        int l=0,r=nums.size()-1;
        while (l>1;
            if (nums[mid] <= nums.back()) r = mid;
            else l=mid+1;
        }
        if (target <= nums.back()) r = nums.size()-1;
        else l=0, r --;
        
        while (l>1;
            if (nums[mid]>=target) r=mid;
            else l=mid+1;
        }
        if (nums[l] == target) return l;
        return -1;
    }
};