LeetCode-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.

经典题。双指针，一个指针先走n步，然后两个同步走，直到第一个走到终点，第二个指针就是需要删除的节点。唯一要注意的就是头节点的处理，比如，
1->2->NULL, n =2; 这时，要删除的就是头节点。

 

/**
 * 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)
            return null;
 
        ListNode fast = head;
        ListNode slow = head;
 
        for(int i=0; i<n; i++){
            fast = fast.next;
        }
 
        //if remove the first node
        if(fast == null){
            head = head.next;
            return head;
        }
 
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
 
        slow.next = slow.next.next;
 
        return head;
    }
}