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.
Solution: head---back------front------>NULL
| |
---> n <----
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(0), *back = &dummy, *front = &dummy;
13 dummy.next = head;
14 while(n--) front = front->next;
15 while(front->next) {
16 front = front->next;
17 back = back->next;
18 }
19 back->next = back->next->next;
20 return dummy.next;
21 }
22 };