description:

一个数列,不知道在哪翻转了一下,现在给定一个值,如果他在这个翻转后的数列里, return 它对应的 index
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).
Note:

Example:

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

answer:

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
            if (nums[mid] < nums[right]) {
                 //较之于普通的二分查找,增加了中轴值得判断
                if (nums[mid] < target && nums[right] >= target) left = mid + 1;
                else right = mid - 1;
            } else {
                if (nums[mid] > target && nums[left] <= target) right = mid - 1;
                else left = mid + 1;
            }
        }
        return -1;
    }
};

relative point get√:

hint :