LeetCode - Rotate Array

Posted 2020-11-07 IncredibleThings

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 1 - Intermediate Array

Space is O(n) and time is O(n). 

class Solution {
    public void rotate(int[] nums, int k) {
        if(nums == null){
            return;
        }
        if(k >= nums.length){
            k = k % nums.length;
        }
        int[] temp = new int[nums.length];
        int count=0;
        for(int i=nums.length-k; i<nums.length; i++){
            temp[count]=nums[i];
            count++;
        }
        for(int i=0; i<nums.length-k; i++){
            temp[count] = nums[i];
            count++;
        }
        System.arraycopy(temp, 0, nums, 0, nums.length);
    }
}

 

Solution 2 - Bubble Rotate

O(1) space,O(n*k) time

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

 

Solution 3 - Reversal

Can we do this in O(1) space and in O(n) time? The following solution does.

Assuming we are given {1,2,3,4,5,6} and order 2. The basic idea is:

1. Divide the array two parts: 1,2,3,4 and 5, 6
2. Reverse first part: 4,3,2,1,5,6
3. Reverse second part: 4,3,2,1,6,5
4. Reverse the whole array: 5,6,1,2,3,4

class Solution {
    public void rotate(int[] nums, int k) {
        if(nums == null){
            return;
        }
        if(k >= nums.length){
            k = k % nums.length;
        }
        reverse(nums, 0, nums.length-k-1);
        reverse(nums, nums.length-k, nums.length-1);
        reverse(nums, 0, nums.length-1);
    }
    
    public void reverse(int[] nums, int left, int right){
        if(nums == null || nums.length == 1){
            return;
        }
        while(left < right){
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}