LeetCode-189 - Rotating Arrays

rotate array

Question Description: Given an array, move the elements of the array k positions to the right, where k is a non-negative number.

Advanced:

• Come up with as many solutions as possible, there are at least three different ways to approach this problem.
• Can you solve this problem using an in-place algorithm with O(1) space complexity?

For example descriptions, please refer to the official website of LeetCode.

Source: LeetCode
Link: https://leetcode-cn.com/problems/rotate-array/
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Solution 1: Array Traversal

First of all, you can know that the number where the index bit i of the array is located should be placed in (i + k) % N , where N is the size of the array, so by traversing the array and declaring a spare array to put the rotated array, The specific process is as follows:

• First move each bit of the array to the position that should be rotated;
• Then reset the nums array.

 public class LeetCode_189 {
    /**
     * 备用空间
     *
     * @param nums
     * @param k
     */
    public static void rotate(int[] nums, int k) {
        int[] copy = new int[nums.length];
        // 首先将数组的每一位移到应该轮转到的位置
        for (int i = 0; i < nums.length; i++) {
            copy[(i + k) % nums.length] = nums[i];
        }

        // 然后重置nums数组
        for (int i = 0; i < nums.length; i++) {
            nums[i] = copy[i];
        }
    }

    public static void main(String[] args) {
        // 测试用例
        int[] nums = new int[]{1, 2, 3, 4, 5, 6, 7};
        System.out.println("轮转之前：");
        for (int num : nums) {
            System.out.print(num + " ");
        }
        System.out.println();

        System.out.println("轮转之后：");
        rotate(nums, 3);
        for (int num : nums) {
            System.out.print(num + " ");
        }
    }
}

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