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
.
思路:先遍历一遍,得出链表长度,注意k可能大于len令k%=len,将尾节点next指针指向首节点,形成一个环 ,然后接着跑len-k步, 从这里断开,就是要求的结果。
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 if(head==NULL || k==0) return head; 13 14 int len=1; 15 ListNode *p=head; 16 while(p->next) 17 { 18 len++; 19 p=p->next; 20 } 21 22 int step=len-k%len; 23 24 p->next=head;//尾节点next指针指向首节点 25 //向后走step步 然后断开 26 27 for(int i=0;i<step;i++) 28 p=p->next; 29 30 head=p->next; 31 p->next=NULL; 32 33 return head; 34 } 35 };