Lintcode174——删除链表中倒数第n个节点

题目：

给定一个链表，删除链表中倒数第n个节点，返回链表的头节点。

又get了一方法

题解：

class ListNode {
int val;
ListNode next;
ListNode(int val) {
this.val = val;
this.next = null;
}
}
public class Solution {
public static ListNode removeNthFromEnd(ListNode head, int n) {
if (n <= 0) {
return null;
}

ListNode dummy = new ListNode(0);
dummy.next = head;

ListNode preDelete = dummy;
for (int i = 0; i < n; i++) {
if (head == null) {
return null;
}
head = head.next;
}
while (head != null) {
head = head.next;
preDelete = preDelete.next;
}
preDelete.next = preDelete.next.next;
return dummy.next;
}
public static void main(String[] args){
ListNode head=new ListNode(1);
ListNode head1=new ListNode(2);
ListNode head2=new ListNode(3);
ListNode head3=new ListNode(4);
ListNode head4=new ListNode(5);
head1.next=head2;
head2.next=head3;
head3.next=head4;
head4.next=null;
removeNthFromEnd(head,2);
System.out.println(head.val+" "+head1.val+" "+head2.val);
}
}

思路

链表从后遍历到第n个———->

a)先用一个节点dummy保存head节点的前一节点，最后返回dummy.next

b)用一个循环让head节点右移n个节点

c)用一和dummy同指向的节点preDelete，和head节点同步移动，直到head移动到null(尾节点.next)

d)用preDelete.next=preDelete.next.next删除该节点