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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;
}
}
LeetCode-Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/incrediblechangshuo/p/5444971.html
