LeetCode-33-Search in Rotated Sorted Array

算法描述：

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

解题思路：时间复杂度限制为O（log n）则首先想到用二分法求解。这道题的要点是先判断哪一部分是递增哪一部分是非递增的，或者都是递增的。然后再分段讨论。

    int search(vector<int>& nums, int target) {
        int low = 0;
        int high = nums.size()-1;
        while(low <= high){
            int mid = low + (high - low) /2;
            if(nums[mid]==target) return mid;
            if(nums[mid] >= nums[low]){
                if(nums[low] <= target && target < nums[mid]) high = mid-1;
                else low = mid+1;
            }else {
                if(nums[mid] < target && target <= nums[high]) low = mid+1;
                else high = mid-1;
            }
        }
        return -1;
    }