Search in rotated Sorted Array II

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What if duplicates is allowed?

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

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

Analysis: When a repeating element is allowed to appear, then the sequence that the search in rotated Sorted array Line14:nums[low] <= nums[mid) can determine is not a monotonically increasing sequence, for example [3,1,3,3,3]. In this case, you simply split the original condition into two scenarios:

1. Nums[low] < Nums[mid] Then the sequence must be incremented, the code is in the search in rotated Sorted array line14~17

2. Nums[low] = = Nums[mid] because it is not able to determine where the number is increased, reduced, after low++ observation. Because low++ after the Nums[low] will be larger or remain unchanged, become larger to meet Nums[low] > Nums[mid], the same will indicate that has entered the repeating number area.

Run Time 11ms

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

Search in rotated Sorted Array II