LeetCode Rotate List

链表右旋算法

最新推荐文章于 2019-05-16 11:19:14 发布

原创最新推荐文章于 2019-05-16 11:19:14 发布 · 259 阅读

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本文介绍了一种链表右旋算法的实现方法，通过计算链表长度并调整指针指向来完成链表的旋转。该算法适用于非负整数k的情况。

题目：

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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
	ListNode *rotateRight(ListNode *head, int k) {
		if (head == NULL)
			return head;
		ListNode *cur = head;
		//s为节点个数
		int s = 1;
		while (cur->next != NULL) {
			s++;
			cur = cur->next;
		}
		ListNode *tail = cur;
		k = (s-k%s)%s;
		if (k == 0)
			return head;
		cur = head;
		int a = 1;
		while (a < k) {
			cur = cur->next;
			a++;
		}
		ListNode *head2 = cur->next;
		cur->next = NULL;
		tail->next = head;
		return head2;
	}
};