Remove Nth Node From End of List - LeetCode

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.

这道题一定要注意的就是，处理头结点要被去除的情况！！

if(count == 0){        return head.next;        }

AC代码如下：

/** * 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) {        if(head == null){            return head;        }                ListNode current = head;        int count = 0;        while(current != null){            count ++;            current = current.next;        }                int len = count;        if(len == 1){            return null;        }        count = len - n;         if(count == 0){        return head.next;        }        current = head;        ListNode pre = head;        for(int i = 0; i < count; i++){            pre = current;            current = current.next;        }        pre.next = current.next;        return head;    }}

更好的解法：

public class Solution {    public 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;    }}