lintcode: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]

class Solution {
private:

    void swap(vector<int> &nums, int p1, int p2)
    {
        int tmp = nums[p1];
        
        nums[p1] = nums[p2];
        nums[p2] = tmp;
    }

    void RotedImpl(vector<int> &nums, int begin, int end) {
        
        if (nums.size() == 0 || begin == end)
            return;
            
        int front = begin;
        int rear  = end;
        
        while (front < rear)
        {
            swap(nums, front, rear);
            front++;
            rear--;
        }
    }
    
public:
    void recoverRotatedSortedArray(vector<int> &nums) {
        // write your code here
        int curIdx = 0;
        
        while (curIdx < nums.size()-1)
        {
            if (nums[curIdx] > nums[curIdx+1])
            {
                break;
            }
            curIdx++;
        }
        
        if (curIdx != nums.size()-1)
        {
            RotedImpl(nums, 0, curIdx);
            RotedImpl(nums, curIdx+1, nums.size()-1);
            RotedImpl(nums, 0, nums.size()-1);
        }
    }
};