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.

解题思路：

要想one pass解决，需用fast 和slow 双指针。

首先先让faster从起始点往后跑n步。然后再让slower和faster一起跑，直到faster==null时候，slower所指向的node就是需要删除的节点。

注意，一般链表删除节点时候，需要维护一个prev指针，指向需要删除节点的上一个节点。

为了方便起见，当让slower和faster同时一起跑时，就不让 faster跑到null了，让他停在上一步，faster.next==null时候，这样slower就正好指向要删除节点的上一个节点，充当了prev指针。这样一来，就很容易做删除操作了。

slower.next = slower.next.next(类似于prev.next = prev.next.next)。

同时，这里还要注意对删除头结点的单独处理，要删除头结点时，没办法帮他维护prev节点，所以当发现要删除的是头结点时，直接让head = head.next并returnhead就够了。

Use fast and slow pointers. The fast pointer is n steps ahead of the slow pointer. When the fast reaches the end, the slow pointer points at the previous element of the target element.

Java code:

/**
* 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 slow = head, fast = head;
// The fast pointer is n steps ahead of the slow pointer.
for(int i =0;i < n; i++) {
fast = fast.next;
}
//remove the head
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;
}
}

Reference:

1. http://www.programcreek.com/2014/05/leetcode-remove-nth-node-from-end-of-list-java/
2. http://www.cnblogs.com/springfor/p/3862219.html