題目:
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.
分析:
先周遊一遍,得對外連結表的長度len,然後把連結清單首尾相連(構成環),然後令K=len-K%len,得出截斷環的位置,移動相應位置後,将環截斷即得到結果。
代碼:
/**
* 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==nullptr||k==)
return head;
int len=;
ListNode *p=head;
while(p->next){
len++;
p=p->next;
}
p->next=head;
k=len-k%len;
for(int i=;i<k;i++){
p=p->next;
}
head=p->next;
p->next=nullptr;
return head;
}
};