33. Search in Rotated Sorted Array

Suppose a sorted array 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.

 

// 时间复杂度 O(log n)  二分查找 先判断mid在左侧还是右侧递增序列 然后二分查找target
class Solution
{
public:
    int search(vector<int>& nums, int target)
    {
        int first = 0;
        int last = first + nums.size(); // 尾后

        while (first != last)
        {
            const int mid = first + (last - first) / 2;

            if (nums[mid] == target) // mid等于target直接返回
            {
                return mid;
            }

            if (nums[first] <= nums[mid]) // mid位于左边的递增序列
            {
                if (nums[first] <= target && target < nums[mid]) // target在first与mid之间
                {
                    last = mid;
                }
                else
                {
                    first = mid + 1;
                }
            }
            else // mid位于右边的递增序列
            {
                if (nums[mid] < target && target <= nums[last - 1])
                {
                    first = mid + 1;
                }
                else
                {
                    last = mid;
                }
            }
        }

        return -1;
    }
};