leetcode 189. Rotate Array(Linkedlist）_linkedlist取尾部-优快云博客

本文链接：https://blog.youkuaiyun.com/DurianPudding/article/details/82813071

本文探讨了如何在原地且使用O(1)额外空间条件下，将数组向右轮换k步的具体算法实现。通过示例展示了轮换过程，并讨论了至少三种不同的解决方案。

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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?

题意：

轮换数组，将尾部数字轮换到首部

思路：

用Linkedlist 循环每次取尾部数字加入到首部

知识点：

void不返回
LinkedList是双向链表，

getLast()
addFirst(element)添加到首部
add(element)顺序添加
removeLast()
get(index)
LinkedList<Integer> A = new LinkedList<Integer>();初始化

bug:

看到void以为要打印输出，但是这个结果是stdout，与answer不同。应该用nums数组存储结果，不用打印

class Solution {
    public void rotate(int[] nums, int k) {
        LinkedList<Integer> A = new LinkedList<Integer>();//初始化
        int len = nums.length;
        for(int i=0;i<nums.length;i++){
            A.add(nums[i]);//nums转换成LinkedList
        }
        for(int i=0;i<k;i++){
            int b = A.getLast();//获取尾部元素
            A.addFirst(b);//将元素添加到首部
            A.removeLast();//删除尾部元素
        }
        for(int i=0;i<len;i++){
            nums[i] = A.get(i);//将LinkedList转换为数组
        }
    }
}