Search in Rotated Sorted Array

 

 

 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.

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

 

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l=0,h=nums.size(),mid;
        int n=h;
        if(nums.size()==0)return -1;
        else if(nums.size()==1)
        {
            if(nums[0]==target)return 0;
            else
                return -1;
        }
        else if(h==2)
        {
            if(target==nums[0])
                return 0;
            else if(target==nums[1])
                return 1;
            else
                return -1;
        }
        int m=-1,first=nums[0],last=nums[n-1];
        --h;
         while(l<=h)
        {
            mid=(l+h)/2;
             if((mid-1)>=0&&nums[mid-1]>nums[mid])
             {
                 m=mid;
                 break;
             }
             else if(mid+1<n&&nums[mid]>nums[mid+1])
             {
                 m=mid+1;
                 break;
             }else if(nums[mid]<first&&nums[mid]<last)
             {
                 h=mid-1;
             }else{
                 l=mid+1;
             }
         }
        
        if(m==-1)
        {
            l=0,h=n-1;
        }else
        if(target>=nums[m]&&target<=last)
        {
            l=m,h=n;
        }else if(target>=first&&target<=nums[m-1])
        {
            l=0,h=m-1;
        }
        else{
            return -1;
        }
        while(l<=h)
        {
            mid=(l+h)/2;
            if(target==nums[mid])
                return mid;
            else if(target<nums[mid])
            {
               h=mid-1;
            }else if(target>nums[mid]){
                l=mid+1;
            }
            
        }
        return -1;
    }
};