39. 恢复旋转排序数组

Description

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

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]

Example

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

Challenge
In-place, O(1) extra space and O(n) time.
-- 要点：三步旋转法

public class Solution {
    /**
     * @param nums: An integer array
     * @return: nothing
     */
    public void recoverRotatedSortedArray(List<Integer> nums) {
        // write your code here
        int index = 0;
        for (int i=0; i<nums.size()-1;i++ ){
            if (nums.get(i) > nums.get(i+1)){
                //得到需要反转的数的关键位置；
                index = i;
                break;
            }
            
        } 
        //判断如果是原数组
        if(index == 0){
            return;
        }

        recover(nums,0,index);
        recover(nums,index+1,nums.size()-1);
        recover(nums,0,nums.size()-1);
            
    } 
    
    
    public void recover(List<Integer> nums,int start, int end){
        int val = end-start;
        
        for (int i=start; i<= (start+ val/2); i++){
        	
            int temp = nums.get(i);
            nums.set(i,nums.get(end-i + start));
            nums.set(end-i+ start,temp);
        } 
        
    }
    
}

说明
什么是旋转数组？

比如，原始数组为[1,2,3,4], 则其旋转数组可以是[1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]
样例
[4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5]

挑战
使用O(1)的额外空间和O(n)时间复杂度