題目：

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

程式碼：

-自己寫的(16ms)：

int search(vector<int>& nums, int target) {

	int len = nums.size();

	if (len == 0)return -1;

	if (target == nums[0])return 0;
	else if (target > nums[0]) {
		for (int i = 0; i < len; i++) {

			if (nums[i] == target)return i;
			if (nums[i] > target)return -1;

		}
	}
	else {
		for (int i = len - 1; i >= 0; i--) {

			if (nums[i] == target)return i;
			if (nums[i] < target)return -1;
			
		}
	}
	return -1;
}
 

-網上的程式碼(8ms):

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;
            else if (nums[mid] < nums[right]) {
                if (nums[mid] < target && nums[right] >= target) left = mid + 1;
                else right = mid - 1;
            } else {
                if (nums[left] <= target && nums[mid] > target) right = mid - 1;
                else left = mid + 1;
            }
        }
        return -1;
    }
 

注：

  複雜度為O(logN)的話，就說明用二分搜尋法