leetcode--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.

算法：双指针

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head==null)
return null;
ListNode p=head;
for(int i=1;i<n;i++){
p=p.next;
}
ListNode s,q;
s=null;
q=head;
while(p.next!=null){
s=q;
q=q.next;
p=p.next;
}
if(s==null)
return q.next;
s.next=q.next;
q.next=null;
q=null;
return head;
}
}

c++：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
if(!head)
return NULL;
ListNode *p,*q,*s;
p=head;
// head已遍历,k从1开始
int k=1;
while(k<n){
p=p->next;
k++;
}
q=head;
s=NULL;
while(p->next){
s=q;
q=q->next;
p=p->next;
}
if(s){
s->next=q->next;
delete q;
return head;
}
// 一个节点,而且n==1
return q->next;
}
};