题目描述:
给定一个链表,删除链表中倒数第n个节点,返回链表的头节点。
注意事项
链表中的节点个数大于等于n
样例
给出链表1->2->3->4->5->null和 n = 2.
删除倒数第二个节点之后,这个链表将变成1->2->3->5->null.
ac代码:
/**
* Definition of ListNode
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: The first node of linked list.
* @param n: An integer.
* @return: The head of linked list.
*/
ListNode *removeNthFromEnd(ListNode *head, int n)
{
// write your code here
ListNode *dummy,*r;
dummy=new ListNode(0);
dummy->next=head;
head=dummy;
//添加头节点。
int sum=0;
while(head!=NULL)
{
sum++;
head=head->next;
}
sum=sum-n;
head=dummy;
while(sum>1)
{
head=head->next;
sum--;
}
head->next=head->next->next;
return dummy->next;
}
};