189. Rotate Array

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?

Solution

Approach 1: Using extra array.

Approach 2: Using cyclic replacement.

Approach 3: Using reverse.

Code (Python)

Approach 1:

Approach 2:

    def rotate(self, array, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        k = k % len(array)
        if k == 0 or len(array) == 1:
            return
        num_switches = 0
        beginning = 0
        index = 0
        number = array[index]
        while True:
            new_index = (index + k) % len(array)
            new_number = array[new_index]
            array[new_index] = number
            #print array
            num_switches += 1
            index = new_index
            number = new_number

            if index == beginning:
                beginning += 1
                index = beginning
                number = array[index]
            
            if num_switches == len(array):
                break

Approach 3:

Code (C++)

Approach 1:

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int numsSize = nums.size();
        int tmp[numsSize];
        for (int i = 0; i < numsSize; ++i)
        {
            tmp[(i + k) % numsSize] = nums[i];
        }
        for (int i = 0; i < numsSize; ++i)
        {
            nums[i] = tmp[i];
        }
    }
};

Approach 2:

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int numsSize = nums.size();
        int count = 0;
        for (int start = 0; count < numsSize; ++start)
        {
            int curr = start;
            int prevNum = nums[start];
            do
            {
                curr = (curr + k) % numsSize;
                int tmp = nums[curr];
                nums[curr] = prevNum;
                prevNum = tmp;
                count++;
            } while (curr != start);
        }
    }
};

Approach 3:

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int numsSize = nums.size();
        if (numsSize == 0) return;
        int realK = k % numsSize;
        reverse(nums, 0, numsSize - 1);
        reverse(nums, 0, realK - 1);
        reverse(nums, realK, numsSize - 1);
    }
private:
    void reverse(vector<int>& nums, int head, int tail) {
        while (head < tail)
        {
            int tmp = nums[head];
            nums[head] = nums[tail];
            nums[tail] = tmp;
            head++;
            tail--;
        }
    }
};