链表题again. 实现链表节点的循环移位. 注意k的边界范围。
思路:
1. 得到链表长度N,指针end指向尾节点
2. 找到新链表头节点在老链表中的前驱节点,记为p. p = N-k%N-1
3. end指向原链表head节点,此时链表为环状。更新链表头节点位置,head指向p->next。将p->next置为NULL,成为尾节点。
Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given1->2->3->4->5->NULL
and k =2
,
return4->5->1->2->3->NULL
.
代码
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *rotateRight(ListNode *head, int k) { 12 // Note: The Solution object is instantiated only once and is reused by each test case. 13 if(head == NULL || k==0) 14 return head; 15 16 ListNode *end = head; 17 int N = 1; 18 while(end->next) 19 { 20 ++N; 21 end = end->next; 22 } 23 int p = N-k%N-1; 24 ListNode *pre = head; 25 if(p<0) 26 return head; 27 else 28 { 29 for(int i=0; i<p; ++i) 30 pre = pre->next; 31 end->next = head; 32 head = pre->next; 33 pre->next = NULL; 34 } 35 return head; 36 } 37 };