README_EN.md

189. Rotate Array

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Description

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

 

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 <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

 

Follow up:

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

Solutions

Python3

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k %= n
        if n < 2 or k == 0:
            return
        nums[:] = nums[::-1]
        nums[:k] = nums[:k][::-1]
        nums[k:] = nums[k:][::-1]

Java

class Solution {
    public void rotate(int[] nums, int k) {
        if (nums == null) {
            return;
        }
        int n = nums.length;
        k %= n;
        if (n < 2 || k == 0) {
            return;
        }

        rotate(nums, 0, n - 1);
        rotate(nums, 0, k - 1);
        rotate(nums, k, n - 1);
    }

    private void rotate(int[] nums, int i, int j) {
        while (i < j) {
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
            ++i;
            --j;
        }
    }
}

JavaScript

the elements in the range k~n-1 of the array to the front with the native API.

/**
 * @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.splice(0, 0, ...nums.splice(-k, k));
};

Use three array reverses implemented by double pointers

/**
 * @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;
    // Use three array reverses
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
};
function reverse(nums, start, end) {
    // reverse implemented by double pointers
    while (start < end) {
        const temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start += 1;
        end -= 1;
    }
}

Go

func rotate(nums []int, k int) {
	n := len(nums)
	k %= n

	reverse(nums, 0, n-1)
	reverse(nums, 0, k-1)
	reverse(nums, k, n-1)
}

func reverse(nums []int, i, j int) {
	for i < j {
		nums[i], nums[j] = nums[j], nums[i]
		i++
		j--
	}
}

Rust

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let n = nums.len();
        let k = k as usize % n;
        if n == 1 || k == 0 {
            return;
        }

        nums.reverse();
        nums[..k].reverse();
        nums[k..].reverse();
    }
}

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