LeetCode189. Rotate Array

Posted 2022-12-04 月盡天明

https://leetcode.com/problems/rotate-array/

Description

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?

---

题目要求空间复杂度为O(1)，没有要求时间复杂度。

---

Solution 1

将数组依次向后移动，一共移动 k 次。

public void rotate(int[] nums, int k) 
        for(int i = 0; i < k; i++) 
            int temp = nums[nums.length - 1];
            for(int j = nums.length - 1; j > 0; j--) 
                nums[j] = nums[j - 1];
            
            nums[0] = temp;
        
    

---

Solution 2

/**
	 * 将后面的往前移动
	 * 74ms
	 * 
	 * @param nums
	 * @param k
	 */
	public void rotate1(int[] nums, int k) 
		if (nums == null || nums.length == 0 || nums.length == 1) 
			return;
		
		if (k > nums.length) 
			k = k % nums.length;
		
		for (int i = 0; i < k; i++) 
			int temp = nums[nums.length - k + i];
			for (int j = nums.length - k + i; j > i; j--) 
				nums[j] = nums[j - 1];
			
			nums[i] = temp;
		
	

---

Solution 3

/**
	 * 数组拷贝
	 * 1ms
	 * 
	 * @param nums
	 * @param k
	 */
	public void rotate2(int[] nums, int k) 
		if (nums == null || nums.length == 0 || nums.length == 1) 
			return;
		
		if (k > nums.length) 
			k = k % nums.length;
		
		int[] result = new int[nums.length];
		for (int i = 0; i < k; i++) 
			result[i] = nums[nums.length - k + i];
		
		for (int i = k; i < nums.length; i++) 
			result[i] = nums[i - k];
		
		System.arraycopy(result, 0, nums, 0, nums.length);
	

---

Solution 4

可以通过递归方式来实现。将数组分为前后两个分别做翻转，然后整体再翻转。

/**
	 * 递归方式，左右两个数组分别倒置，然后整体再倒置
	 * 1ms
	 * 
	 * @param nums
	 * @param k
	 */
	public void rotate3(int[] nums, int k) 
		if (nums == null || nums.length == 0 || nums.length == 1) 
			return;
		
		if (k > nums.length) 
			k = k % nums.length;
		
		reverse(nums, 0, nums.length - k - 1);
		reverse(nums, nums.length - k, nums.length - 1);
		reverse(nums, 0, nums.length - 1);
	

	public void reverse(int[] arr, int left, int right) 
		if (arr == null || arr.length == 0 || arr.length == 1) 
			return;
		
		int temp = 0;
		while (left < right) 
			temp = arr[left];
			arr[left] = arr[right];
			arr[right] = temp;
			left++;
			right--;