leetcode 61. Rotate List (链表循环移动 k 位)
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
- 链表操作
- 向右循环移动 位
- 注意一些 corner case
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(!k)return head;
if(head==NULL||head->next==NULL)return head;
ListNode* p=head;
ListNode* pp=head;
int len=1;
while(p->next!=NULL)p=p->next,len++;
k=k%len;
p=head;
for(int i=0;i<k;i++)
p=p->next;
while(p->next!=NULL){
p=p->next;
pp=pp->next;
}
p->next=head;
ListNode* res=pp->next;
pp->next=NULL;
return res;
}
};