Rotate List - LeetCode

最新推荐文章于 2019-08-30 09:58:00 发布

GigaCat

最新推荐文章于 2019-08-30 09:58:00 发布

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分类专栏： LeetCode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/gigacat/article/details/19800917

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本文介绍了一种链表操作——右旋，并提供了一段实现代码。该操作将链表向右旋转k个位置，通过示例展示了操作前后的链表变化。文章还提供了另一段参考代码，进一步解释了实现细节。

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Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

没有完全AC，还有细节小问题，但是大体逻辑是这样的。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
       if(head == null){
	            return null;
	        }
		 if(n == 0)
		        return head;
	        ListNode newhead = new ListNode(0);
	        int count = 1;
	        ListNode current = new ListNode(0);
	        current = head;
	        
	        while(current.next != null){
	            count ++;
	            current = current.next;
	        }
	      //  System.out.println(count);
	        if(count == 1)
		        return head;
		    if(n > count)
		    n = count;
	        ListNode tail = current;
	        current = head;
	        
	        for(int i = 1; i < count - n; i++){
	            current = current.next;
	        }
	        newhead.next = current.next;
	        tail.next = head;
	        current.next = null;
	          return newhead.next;
    }
    
}

参考的AC代码：

public class Solution {
    private int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            length ++;
            head = head.next;
        }
        return length;
    }
    
    public ListNode rotateRight(ListNode head, int n) {
        if (head == null) {
            return null;
        }
        
        int length = getLength(head);
        n = n % length;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        ListNode tail = dummy;
        for (int i = 0; i < n; i++) {
            head = head.next;
        }
        
        while (head.next != null) {
            tail = tail.next;
            head = head.next;
        }
        
        head.next = dummy.next;
        dummy.next = tail.next;
        tail.next = null;
        return dummy.next;
    }
}