61. Rotate List(M);19. Remove Nth Node From End of List(M)

61. Rotate List(M)

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.

Total Accepted: 102574
Total Submissions: 423333
Difficulty: Medium

 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 == nullptr || k == 0) return head;
13         int len = 1;
14         ListNode* p = head;
15         while (p->next) { //
16             len++;
17             p = p->next;
18         }
19         k = len - k % len;
20         p->next = head; //
21         for(int step = 0; step < k; step++) {
22             p = p->next; //
23         }
24         head = p->next; //
25         p->next = nullptr; //
26         return head;
27     }
28 };

16ms 19.35%

 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 {// by guxuanqing@gmail.com
10 public:
11     ListNode* rotateRight(ListNode* head, int k)
12     {
13         if(NULL == head || NULL == head->next) return head;//null or one node
14         ListNode *p = NULL, *q = head;
15         int len = 1;
16         while (q->next)//cal the length of the list
17         {
18            ++len;
19             q = q->next;
20         }
21         q->next = head;
22         k %= len;
23         int tmp = len - k;
24         q = head;
25         while (--tmp)
26         {
27             q = q->next;
28         }
29         head = q->next;
30         q->next = NULL;
31         return head;
32     }
33 };

19. Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.

Total Accepted: 169535
Total Submissions: 517203
Difficulty: Medium

 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 *removeNthFromEnd(ListNode *head, int n) {
12         ListNode dummy(-1);
13         dummy.next = head;
14         ListNode *p = &dummy, *q = &dummy;
15         for (int i = 0; i < n; i++)//q先走n步
16             q = q->next;
17         while(q->next) {
18             p = p->next;
19             q = q->next;
20         }
21         ListNode *tmp = p->next;
22         p->next = p->next->next;
23         delete tmp;
24         return dummy.next;
25     }
26 };

6ms 64.75%

 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 {// by guxuanqing@gmail.com
10 public:
11     ListNode* removeNthFromEnd(ListNode* head, int n) {
12         if(NULL == head || NULL == head->next) return NULL;//null or one node
13         ListNode dummy(-1);
14         dummy.next = head;
15         ListNode *p = &dummy, *q = &dummy;
16         int tmp = n;
17         while (tmp--)
18         {
19             q = q->next;
20         }
21         while (q->next)
22         {
23             q = q->next;
24             p = p->next;
25         }
26         ListNode *rnode = p->next;
27         p->next = rnode->next;
28         delete rnode;
29         return dummy.next;
30     }
31 };