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Question

Formatted question description: https://leetcode.ca/all/189.html

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?

Algorithm

O(1) extra space.

Similar to the method of flipping characters, the idea is to

  1. flip the first n-k numbers first,
  2. then flip the last k numbers,
  3. and finally flip the entire array.

Proces as below:

[1,2,3,4, 5,6,7] => [4,3,2,1, 5,6,7]

[4,3,2,1, 5,6,7] => [4,3,2,1, 7,6,5]

[4,3,2,1, 7,6,5] => [5,6,7, 1,2,3,4]

Easier implemetatioin is:

  1. and finally flip the entire array first,
  2. then flip the first k numbers,
  3. flip the last n-k numbers

Proces as below:

[1,2,3,4, 5,6,7] => [7,6,5, 4,3,2,1]

[7,6,5, 4,3,2,1] => [5,6,7, 4,3,2,1]

[5,6,7, 4,3,2,1] => [5,6,7, 1,2,3,4]

Code

  • import org.apache.commons.lang3.ArrayUtils;

public class Rotate_Array {

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

            int n = nums.length;

            // [1,2,3,4,5,6,7] => [4,3,2,1,5,6,7]
            ArrayUtils.reverse(nums, 0, n - k); // @note: org.apache.commons.lang3.ArrayUtils

            // [4,3,2,1,5,6,7] => [4,3,2,1,7,6,5]
            ArrayUtils.reverse(nums, n - k, nums.length);

            // [4,3,2,1,7,6,5] => [5,6,7,1,2,3,4]
            ArrayUtils.reverse(nums, 0, nums.length);
        }
    }

    class Solution_kTimesAllItemsMoving {
        public void rotate(int[] nums, int k) {

            while ( k > 0) {
                k--;

                int tmp = nums[nums.length - 1]; // last one
                for (int i = nums.length - 1; i >= 1; i--) {
                    nums[i] = nums[i - 1];
                }

                nums[0] = tmp;
            }
        }
    }
}

############

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;
        }
    }
}

  • // OJ: https://leetcode.com/problems/rotate-array/
// Time: O(N)
// Space: O(K)
class Solution {
public:
    void rotate(vector<int>& A, int k) {
        int N = A.size();
        k %= N;
        if (k == 0) return;
        vector<int> tmp(k);
        for (int i = 0; i < k; ++i) tmp[i] = A[N + i - k];
        for (int i = N - k - 1; i >= 0; --i) A[i + k] = A[i];
        for (int i = 0; i < k; ++i) A[i] = tmp[i];
    }
};

  • 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]

############

class Solution(object):
  def rotate(self, nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: void Do not return anything, modify nums in-place instead.
    """
    if len(nums) == 0 or k == 0:
      return

    def reverse(start, end, s):
      while start < end:
        s[start], s[end] = s[end], s[start]
        start += 1
        end -= 1

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


  • 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--
	}
}


  • /**
 * @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));
};


  • using System;

public class Solution {
    public void Rotate(int[] nums, int k) {
        if (nums.Length == 0 || k % nums.Length == 0) return;
        k = k % nums.Length;
        Algo2(nums, k);
    }
    
    private void Algo1(int[] nums, int k)
    {
        var copy = new int[nums.Length];
        Array.Copy(nums, copy, nums.Length);
        var j = nums.Length - k;
        for (var i = 0; i < nums.Length; ++i)
        {
            if (j == nums.Length) j = 0;
            nums[i] = copy[j];
            ++j;
        }
    }
    
    private void Algo2(int[] nums, int k)
    {
        var gcd = Gcd(nums.Length, k);
        var count = nums.Length / gcd;
        for (var i = 0; i < gcd; ++i)
        {
            var p = i;
            var first = nums[p]; 
            for (var j = 0; j + 1 < count; ++j)
            {
                var q = p - k;
                if (q < 0) q += nums.Length;
                nums[p] = nums[q];
                p = q;
            }
            nums[i + k] = first;
        }
    }
    
    private int Gcd(int m, int n)
    {
        while (n > 0)
        {
            var x = m % n;
            m = n;
            n = x;
        }
        return m;
    }
}

  • 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();
    }
}


  • /**
 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);
}

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