[LeetCode] Rotate Array

最新推荐文章于 2024-01-03 12:46:00 发布

weixin_30341745

最新推荐文章于 2024-01-03 12:46:00 发布

阅读量55

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zgcy123456/p/4300366.html

本文介绍了一种在原地将数组元素向右旋转k步的方法，并提供了一个具体的Java实现示例。该算法通过使用两个临时变量来交换数组中的元素，从而达到在O(1)额外空间复杂度下完成旋转的目标。

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].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

思路：两个临时变量，一个存放需要写入的值，一个存放取出的值。

public class RotateArray {

	public static void main(String[] args) {
		int[] temp = { 1, 2, 3, 4, 5, 6, 7 };
		rotate(temp, 3);
		for (int i = 0; i < temp.length; i++) {
			System.out.print(temp[i] + "  ");
		}
	}

	public static void rotate(int[] nums, int k) {
		if (null == nums) {
			return;
		}
		int length = nums.length;
		for (int i = 0; i < k; i++) {
			// 需要写入的值
			int input = nums[0];
			// 需要取出来的值
			int next = nums[length - 1];
			for (int j = 1; j < length; j++) {
				next = nums[j];
				nums[j] = input;
				input = next;
			}
			nums[0] = next;
		}
	}
}

转载于:https://www.cnblogs.com/zgcy123456/p/4300366.html