19.2.9 [LeetCode 61] Rotate List

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

Example 1:

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

题意

把链表循环右移k个

题解

 1 class Solution {
 2 public:
 3     ListNode* rotateRight(ListNode* head, int k) {
 4         if (head == NULL)return NULL;
 5         ListNode*pre, *p = head;
 6         int cnt = 0;
 7         while (p != NULL) {
 8             p = p->next;
 9             cnt++;
10         }
11         k %= cnt;
12         if (k == 0)return head;
13         p = head;
14         while (k--)
15             p = p->next;
16         pre = head;
17         while (p->next != NULL) {
18             p = p->next;
19             pre = pre->next;
20         }
21         ListNode*ans = pre->next;
22         pre->next = NULL;
23         p->next = head;
24         return ans;
25     }
26 };

考虑空链和不移动的特殊情况

原文地址：https://www.cnblogs.com/yalphait/p/10357610.html