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?

 

Approach #1: Extra array. [Java]

class Solution {
    public void rotate1(int[] nums, int k) {
        int n = nums.length;
        int[] ans = new int[n];
        int start = 0, next = 0;
        while (n--) {
            start = next;
            next = (start + k) % n;
            nums[next] = nums[start];
        }
        for (int i = 0; i < n; ++i) {
            int rev = (i + k) % n;
            ans[rev] = nums[i];
        }
        for (int i = 0; i < n; ++i)
            nums[i] = ans[i];
        return ;
    }
}

　　

Approach #2: Reverse. [Java]

public class Solution {
    public void rotate(int[] nums, int k) {
        k %= nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
    }
    public void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

　　

Approach #3: Cyclic Replacement. [Java]

public class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        int count = 0;
        for (int start = 0; count < nums.length; start++) {
            int current = start;
            int prev = nums[start];
            do {
                int next = (current + k) % nums.length;
                int temp = nums[next];
                nums[next] = prev;
                prev = temp;
                current = next;
                count++;
            } while (start != current);
        }
    }
}