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[LeetCode] 189. Rotate Array

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

Follow up:

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?

Example 1:

Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]

Constraints:

1 <= nums.length <= 2 * 10^4
It's guaranteed that nums[i] fits in a 32 bit-signed integer.
k >= 0

旋转数组。

题意跟61题几乎一样，唯一不同的是这题的 input 是数组，所以代码上没什么参考性。这个题目的思路很巧妙，叫做三步反转法。我就拿第一个例子遍历好了。数组长度为7，要往右 rotate 三步。三步反转法的思路是

start, [1, 2, 3, 4, 5, 6, 7]

整个rotate，[7, 6, 5, 4, 3, 2, 1]

rotate前k个， [5, 6, 7, 4, 3, 2, 1]

rotate后n - k个，[5, 6, 7, 1, 2, 3, 4]

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} k
 4  * @return {void} Do not return anything, modify nums in-place instead.
 5  */
 6 var rotate = function(nums, k) {
 7     k = k % nums.length;
 8     reverse(nums, 0, nums.length - 1);
 9     reverse(nums, 0, k - 1);
10     reverse(nums, k, nums.length - 1);
11 };
12 
13 var reverse = function(nums, start, end) {
14     while (start < end) {
15         let temp = nums[start];
16         nums[start] = nums[end];
17         nums[end] = temp;
18         start++;
19         end--;
20     }
21 };

Java实现

 1 class Solution {
 2     public void rotate(int[] nums, int k) {
 3         k = k % nums.length;
 4         reverse(nums, 0, nums.length - 1);
 5         reverse(nums, 0, k - 1);
 6         reverse(nums, k, nums.length - 1);
 7     }
 8 
 9     private void reverse(int[] nums, int start, int end) {
10         while (start < end) {
11             int temp = nums[start];
12             nums[start] = nums[end];
13             nums[end] = temp;
14             start++;
15             end--;
16         }
17     }
18 }

2020年1月24日更新

还有一种JS的思路是 pop 再 shift，利用的是 JS 对 array 的各项操作。

时间O(k)

空间O(n)

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} k
 4  * @return {void} Do not return anything, modify nums in-place instead.
 5  */
 6 var rotate = function(nums, k) {
 7     k = k % nums.length;
 8     for (var i = 1; i <= k; i++) {
 9         nums.unshift(nums.pop());
10     }
11 };