19. Remove Nth Node From End of List

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法只需遍历一次链表，通过两个指针的方法定位待删除节点并进行删除操作。

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.

思路：先扫描一遍链表得出链表的总长度len，然后n=n%len。然后再遍历链表到第len-n个节点，让它的next=next.next就行

代码如下(已通过leetcode)

public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode p = head;
ListNode q = head;
int count = 0;
int length=0;
while (p != null) {
if (count == n + 1) {
p = p.next;
q = q.next;
} else {
p = p.next;
count++;
}
length++;
}

System.out.print(q.val);
if(length==1) return null;
if(length==n) head=head.next;
if (q.next.next == null)
q.next = null;
else
q.next = q.next.next;

return head;
}
}