LeetCode Remove Nth Node From End of List

题目

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.

思想：保持前后两个指针，使其距离差为n-1，前者到头时后者即为要删除的元素。

代码：

/**
* 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) {
ListNode *p1=head,*p2=head,*pre=NULL;	//前、后指针，后指针的前驱
for(int i=1;i<n;i++)
if(p1->next!=NULL)
p1=p1->next;
else
break;
while(p1->next!=NULL)
{
pre=p2;
p2=p2->next;
p1=p1->next;
}
if(pre==NULL)
head=p2->next;
else
pre->next=p2->next;
delete p2;
return head;
}
};