[LeetCode 189] Rotate Array (Easy)

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Solution： 三步翻转法

class Solution {
    public void rotate(int[] nums, int k) {
        //三次翻转法
        if (nums == null || nums.length == 0) {
            return;
        }
        
        if (k > nums.length) {
            k = k % nums.length;
        }
        
        int diff = nums.length - k;
        
        if (diff < 0)
            return;
        
        // 3 steps reverse
        reverse (nums, 0, diff - 1);  // reverse first part: 0 ~ (length - k)
        //System.out.println (Arrays.toString (nums));
        
        reverse (nums, diff, nums.length - 1); // reverse second part: (length - k) ~ length
        //System.out.println (Arrays.toString (nums));
        
        reverse (nums, 0, nums.length - 1); // reverse entire array 
        //System.out.println (Arrays.toString (nums));
    }
    
    public void reverse (int[] nums, int start, int end) {
        if (start >= end) {
            return;
        }
        
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            
            start ++;
            end --;
        }
        
        return;
    }
}