0189. Rotate Array (E)

Rotate Array (E)

题目

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?

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题意

将给定数列的指定后半部分与前半部分换位，得到新数组。

思路

最经典的(O(1))空间方法是翻转法：先将左子数组翻转，再将右子数组翻转，最后将整个数组翻转，得到的就是目标数组。

比较直接的是按照步骤一个一个移动元素，或者使用额外数组先将右子数组保存下来再处理。

官方解答还提供了一种循环替换法：以一个元素为起点，直接将该元素放在它应在的位置上，并将该位置上原本的元素继续按上述操作放在下一个位置上，直到回到起点完成一次循环，接着更换起点重复操作即可。当这种放置进行了n次后，所有元素都已经在它应在的位置上。该方法是对暴力法的一种优化。

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代码实现

Java

翻转法

class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        reverse(nums, 0, nums.length - 1 - k);
        reverse(nums, nums.length - k, nums.length - 1);
        reverse(nums, 0, nums.length - 1);
    }

    private void reverse(int[] nums, int left, int right) {
        while (left < right) {
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}

暴力法

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

额外数组

class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        int[] temp = Arrays.copyOfRange(nums, nums.length - k, nums.length);
        for (int i = nums.length - 1; i >= k; i--) {
            nums[i] = nums[i - k];
        }
        for (int i = 0; i < k; i++) {
            nums[i] = temp[i];
        }
    }
}

循环替换

class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        int count = 0;
        
        // 当count==nums.length时，说明所有元素都已经在它应在的位置上
        for (int i = 0; count < nums.length; i++) {
            int j = i;
            int pre = nums[j];
            do {
                int next = (j + k) % nums.length;
                int temp = nums[next];
                nums[next] = pre;
                pre = temp;
                j = next;
                count++;
            } while (j != i);	// 回到起点，说明一次循环完成
        }
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function (nums, k) {
  k %= nums.length
  nums
    .slice(0, nums.length - k)
    .reverse()
    .concat(nums.slice(nums.length - k).reverse())
    .reverse()
    .forEach((v, i) => (nums[i] = v))
}