[leetcode] 【链表】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.

题意

删除链表倒数第n个节点。

题解

两个指针，一个先走n补，然后两个一起走，那么等第一个指针到尾的时候，第二个指针正好在倒数第n个位置。

/**
* 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 newhead(-1);
newhead.next=head;
ListNode *fast=&newhead,*slow=&newhead;
for(int i=0;i<n;i++)
fast=fast->next;
while(fast->next)
{
fast=fast->next;
slow=slow->next;
}
ListNode *temp=slow->next;
slow->next=slow->next->next;
delete temp;
return newhead.next;

}
};