LeetCode[19]Remove Nth Node From End of List 删除链表倒数第n个元素

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.

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思路：设置两个指针a跟b。a指针先移动n步，若此时a指针为空，则表示要删除的是头节点，此时直接返回head->next即可。如果a指针不为空，则将两个指针一起移动，直到a指针指向最后一个节点，令b->next=b->next->next即可删除第你n个节点。

ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head == NULL || head->next == NULL)
return NULL;
ListNode *a=head,*b=head;
for(int i = 0;i < n;i++)
a=a->next;
if(a==NULL)
return head->next;
while(a->next!=NULL)
{
a=a->next;
b=b->next;
}
b->next=b->next->next;
return head;
}