Search in rotated Sorted Array

Suppose a sorted array is rotated on 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 is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

Analysis: Using two-point search method. Mid = (low + high)/2. Consider two scenarios such as:

1. If Nums[low] <= Nums[mid] indicates that the interval is incremented, if the value of target is between the two, then the upper bound is mid-1, otherwise the lower bound is mid + 1;

2. If Nums[low] > Nums[mid] indicates that mid is after the maximum value (example: [4, 5, 6, 0, 1, 2, 3]. Number 0 after 6) If Target is between Nums[mid] and Nums[high], the lower bound is mid + 1; otherwise the upper bound is mid-1.

Run Time 7ms

1 classSolution {2  Public:3     intSearch (vector<int>& Nums,inttarget) {4         if(nums.size () = =0)return-1;5         if(nums.size () = =1){6             if(nums[0] = = target)return 0;7             Else return-1;8         }9         Ten         intLow =0, high = Nums.size ()-1; One          while(Low <=High ) { A             intMid = (low + high)/2; -             if(Nums[mid] = = target)returnmid; -             if(Nums[low] <=Nums[mid]) { the                 if(Nums[low] <= target && target < Nums[mid]) high =mid; -                 ElseLow = mid +1; -             } -             Else{ +                 if(Nums[mid] < target && target <= nums[high]) Low = mid +1; -                 ElseHigh =mid; +             } A         } at         return-1; -     } -};

Search in rotated Sorted Array