在一个递增数组的rotate变换中找target

Search in Rotated Sorted Array

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.

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

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

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