Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
首先遍历链表，得到链表长度。定义两个指针，一前一后，一同遍历链表，直到先出发的指针定位到目标结点，然后进行删除操作。需要对删除头节点的情况进行特殊处理。
/** * 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*p=head; ListNode*q=head; int counter=0; while(p!=NULL) { p=p->next; counter++; } p=head; if(counter-n==0) { head=head->next; delete p; return head; } for(int i=0;p!=NULL&&i<counter-n;i++) { p=p->next; if(i>0) q=q->next; } if(p!=NULL) { q->next=p->next; delete p; } return head; } };
【LeetCode】26.Linked List —Remove Nth Node From End of List 从列表末尾删除第n个节点
原文：https://www.cnblogs.com/hu-19941213/p/11255477.html