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;
}
};