Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given
return
» Solve this problemGiven
1->2->3->4->5->NULL
and k = 2
,return
4->5->1->2->3->NULL
.[解题思路]
首先从head开始跑,直到最后一个节点,这时可以得出链表长度len。然后将尾指针指向头指针,将整个圈连起来,接着往前跑len – k%len,从这里断开,就是要求的结果了。
[Code]
1: ListNode *rotateRight(ListNode *head, int k) {
2: // Start typing your C/C++ solution below
3: // DO NOT write int main() function
4: if(head == NULL || k ==0) return head;
5: int len =1;
6: ListNode* p = head,*pre;
7: while(p->next!=NULL)
8: {
9: p = p->next;
10: len++;
11: }
12: k = len-k%len;
13: p->next = head;
14: int step =0;
15: while(step< k)
16: {
17: p = p->next;
18: step++;
19: }
20: head = p->next;
21: p->next = NULL;
22: return head;
23: }
[Note]
注意K大于len的可能。
注意K大于len的可能。