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:
先令一個指針node跑到最後的一個節點(非NULL)處;
讓node的下一個指針指向頭結點形成環形單鏈表;
然後就是計算要遍歷的次數。這裏分爲三種情況:
1.k>n: k=k%n>0;
2.k
/**
* 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) return NULL;
ListNode* node=head;
int n=1;//這裏必須爲1,否則後面k=k%n的時候不對
while(node->next){
node=node->next;
n++;
}
node->next=head;
if(k %= n)
{
int n1=n-k;//循環次數
while(n1--!=0) node=node->next;
}
ListNode* newhead=head;
newhead=node->next;
node->next=NULL;
return newhead;
}
};