LeetCode 133 Remove Nth Node From End of List

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.

分析：

快慢指针法，让快指针先走n，再一起走。

小技巧：链表问题，新建一个dummy头总是很有帮助的。

/** * 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 null;        ListNode dummy = new ListNode(0);        dummy.next = head;        //快慢指针法        ListNode fast = dummy;        ListNode slow = dummy;        while(n>0){            fast = fast.next;            n--;        }        while(fast.next != null){            fast = fast.next;            slow = slow.next;        }        slow.next = slow.next.next;        return dummy.next;    }}