[lintcode easy]Recover rotated sorted array

Recover Rotated Sorted Array

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

Example

[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 is [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]

////First, find out the index of the greatest number in the array,which means at this point ,it will greater than the next number.

///Second, partition the array to two part from the index

///Reverse them in partition.

///Reverse the whole array.

public class Solution {
    /**
     * @param nums: The rotated sorted array
     * @return: void
     */
    public void recoverRotatedSortedArray(ArrayList<Integer> nums) {
        // write your code
        int n=nums.size();
        int j=0;
        int k=j;
        for(int i=0;i<n-1;i++)
        {
            if(nums.get(i)>nums.get(i+1))
            {
                reverse(nums,0,i);
                reverse(nums,i+1,n-1);
                reverse(nums,0,n-1);
            }
        }
    }

    public void reverse(ArrayList<Integer> nums, int start, int end)
    {
        int temp;
        while(start<end)
        {
            temp=nums.get(start);
            nums.set(start,nums.get(end));
            nums.set(end,temp);
            start++;
            end--;
        }
    }
}