Recover Rotated Sorted Array

Given a rotated sorted array, recover it to sorted array in-place.

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[4, 5, 1, 2, 3]-[1, 2, 3, 4, 5]

Challenge

In-place, O (1) extra space and O (n) time.

Clarification

What is rotated array?

• For example, the orginal array are [1,2,3,4], the rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]

1      Public voidRecoverrotatedsortedarray (arraylist<integer>nums)2     {3         //Write your code4         if(Nums.size () ==0| | Nums.size () ==1)return;5         intI =0;6          for(i=0; i<nums.size () -1;i++)7         {8             if(Nums.get (i) >nums.get (i+1)) Break;9         }Ten         if(i = = Nums.size ()-1)return; OneReverse (nums,0, i); AReverse (Nums,i+1,nums.size ()-1); -Reverse (Nums,0,nums.size ()-1); -         return; the          -          -     } -      +      -      Public voidReverse (arraylist<integer> nums,intLeftintRight ) +     { A          while(left<Right ) at         { -             inttemp =Nums.get (left); - Nums.set (Left,nums.get (right)); - Nums.set (right,temp); -left++; -right--; in          -         } to        +         return; -}

Recover Rotated Sorted Array