189. Rotate Array

Description

Given an integer array nums, 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

Solution 1: Reverse three times

We can assume the length of the array is $n$ and calculate the actual number of steps needed by taking the module of $k$ and $n$, which is $k \bmod n$.

Next, let us reverse three times to get the final result:

Reverse the entire array.
Reverse the first $k$ elements.
Reverse the last $n - k$ elements.

For example, for the array $[1, 2, 3, 4, 5, 6, 7]$, $k = 3$, $n = 7$, $k \bmod n = 3$.

In the first reverse, reverse the entire array. We get $[7, 6, 5, 4, 3, 2, 1]$.
In the second reverse, reverse the first $k$ elements. We get $[5, 6, 7, 4, 3, 2, 1]$.
In the third reverse, reverse the last $n - k$ elements. We get $[5, 6, 7, 1, 2, 3, 4]$, which is the final result.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        def reverse(i: int, j: int):
            while i < j:
                nums[i], nums[j] = nums[j], nums[i]
                i, j = i + 1, j - 1

        n = len(nums)
        k %= n
        reverse(0, n - 1)
        reverse(0, k - 1)
        reverse(k, n - 1)

class Solution {
    private int[] nums;

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

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

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        k %= n;
        reverse(nums.begin(), nums.end());
        reverse(nums.begin(), nums.begin() + k);
        reverse(nums.begin() + k, nums.end());
    }
};

func rotate(nums []int, k int) {
	n := len(nums)
	k %= n
	reverse := func(i, j int) {
		for ; i < j; i, j = i+1, j-1 {
			nums[i], nums[j] = nums[j], nums[i]
		}
	}
	reverse(0, n-1)
	reverse(0, k-1)
	reverse(k, n-1)
}

/**
 Do not return anything, modify nums in-place instead.
 */
function rotate(nums: number[], k: number): void {
    const n: number = nums.length;
    k %= n;
    const reverse = (i: number, j: number): void => {
        for (; i < j; ++i, --j) {
            const t: number = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    };
    reverse(0, n - 1);
    reverse(0, k - 1);
    reverse(k, n - 1);
}

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let n = nums.len();
        let k = (k as usize) % n;
        nums.reverse();
        nums[..k].reverse();
        nums[k..].reverse();
    }
}

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function (nums, k) {
    const n = nums.length;
    k %= n;
    const reverse = (i, j) => {
        for (; i < j; ++i, --j) {
            [nums[i], nums[j]] = [nums[j], nums[i]];
        }
    };
    reverse(0, n - 1);
    reverse(0, k - 1);
    reverse(k, n - 1);
};

public class Solution {
    private int[] nums;

    public void Rotate(int[] nums, int k) {
        this.nums = nums;
        int n = nums.Length;
        k %= n;
        reverse(0, n - 1);
        reverse(0, k - 1);
        reverse(k, n - 1);
    }

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

Solution 2

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        k %= len(nums)
        nums[:] = nums[-k:] + nums[:-k]

doocs / leetcode

189. Rotate Array

Description

Solutions

Solution 1: Reverse three times

Solution 2

简介

发行版

贡献者

近期动态

doocs / leetcode .gitee-modal { width: 500px !important; }

189. Rotate Array

Description

Solutions

Solution 1: Reverse three times

Solution 2

简介

发行版

开源评估指数源自 OSS-Compass 评估体系，评估体系围绕以下三个维度对项目展开评估：

贡献者

近期动态

搜索帮助

doocs / leetcode