[算法]Rotate Array

You may have been using Java for a while. Do you think a simple Java array question can be a challenge? Let's use the following problem to test.

Problem: Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. How many different ways do you know to solve this problem?

Solution 1 - Intermediate Array

In a straightforward way, we can create a new array and then copy elements to the new array. Then change the original array by using System.arraycopy().

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

    if(k > nums.length)

        k=k%nums.length;


    int[] result = new int[nums.length];


    for(int i=0; i < k; i++){

        result[i] = nums[nums.length-k+i];

    }


    int j=0;

    for(int i=k; i<nums.length; i++){

        result[i] = nums[j];

        j++;

    }


    System.arraycopy( result, 0, nums, 0, nums.length );

}

Space is O(n) and time is O(n). You can check out the difference between System.arraycopy() and Arrays.copyOf().

Solution 2 - Bubble Rotate

Can we do this in O(1) space?

This solution is like a bubble sort.

public static void rotate(int[] arr, int order) {

	if (arr == null || order < 0) {

	    throw new IllegalArgumentException("Illegal argument!");

	}


	for (int i = 0; i < order; i++) {

		for (int j = arr.length - 1; j > 0; j--) {

			int temp = arr[j];

			arr[j] = arr[j - 1];

			arr[j - 1] = temp;

		}

	}

}

However, the time is O(n*k).

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. Rotate first part: 4,3,2,1,5,6

3. Rotate second part: 4,3,2,1,6,5

4. Rotate the whole array: 5,6,1,2,3,4

public static void rotate(int[] arr, int order) {

	order = order % arr.length;


	if (arr == null || order < 0) {

		throw new IllegalArgumentException("Illegal argument!");

	}


	//length of first part

	int a = arr.length - order;


	reverse(arr, 0, a-1);

	reverse(arr, a, arr.length-1);

	reverse(arr, 0, arr.length-1);


}


public static void reverse(int[] arr, int left, int right){

	if(arr == null || arr.length == 1)

		return;


	while(left < right){

		int temp = arr[left];

		arr[left] = arr[right];

		arr[right] = temp;

		left++;

		right--;

	}

}

巴特西

[算法]Rotate Array

Solution 1 - Intermediate Array

Solution 2 - Bubble Rotate

Solution 3 – Reversal

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