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.
解法：从后面往前数的第n个
pointerA先从前开始数n个，然后另一个pointerB从开头数，当A到达最后的时候,b就到了合正确的地方
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode record=new ListNode(0); record.next=head; ListNode node=record; for(int i=0;i<n;i++){ if(head==null){ return null; } head=head.next; } while(head!=null){ head=head.next; node=node.next; } node.next=node.next.next; return record.next; } }
leetcode Remove Nth Node From End of List
原文：http://www.cnblogs.com/lilyfindjobs/p/4075317.html