19. Remove Nth Node From End of List
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.
题意：删除链表的倒数第n个节点
代码如下：
/** * Definition for singly-linked list. * function ListNode(val) { * ????this.val = val; * ????this.next = null; * } *//** * @param {ListNode} head * @param {number} n * @return {ListNode} */var removeNthFromEnd = function(head, n) { ????if(head.next==null) return null; ???????//定义两个指针 ???????let pre=head,cur=head; ???????//让cur沿着head向后移动n个节点 ???????for(let i=0;i<n;i++){ ???????????cur=cur.next; ???????} ???????if(!cur) return head.next; ???????//直至cur指向head末端 ???????while(cur.next){ ???????????cur=cur.next; ???????????pre=pre.next; ???????} ???????pre.next=pre.next.next; ???????return head; };19. Remove Nth Node From End of List（js）
原文地址：https://www.cnblogs.com/xingguozhiming/p/10387351.html