leetcode 19:Remove Nth Node From End of List

题目描述

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.

题目大意：删除链表的倒数第N个结点。要求只遍历一次。

解题思路

用两个指针，第一个指针先走N步，第二个指针停在头结点，然后一起走，当第一个指针走到最后时，第二个指针指向要删除结点的前一个结点

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head == null || n == 0)
return null;
ListNode start = new ListNode(0);
start.next = head;
ListNode pHead = start;
ListNode pBehind = start;
for(int i =0 ; i< n;i++){
pHead = pHead.next;
}
while(pHead.next!=null){
pHead=pHead.next;
pBehind = pBehind.next;
}
pBehind.next = pBehind.next.next;
return start.next;
}
}