LeetCode Solutions : Remove Nth Node From End of List

【題目描述】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.

【算法思路】這道題的關鍵在於找倒數第n個結點的前驅結點：可以設置慢行指針slow和快行指針fast，讓快行指針fast先走n步，然後同步地移動慢行指針slow和快行指針fast，這樣它們之間的距離永遠相差n個結點，當fast指針移動最後一個結點的時候，slow指向倒數第n個結點；

【編程步驟】

 * 1. 處理特殊情況：如果被移除的結點爲頭結點，則移除，並返回第二個結點作爲頭結點；
 * 2. 如果爲其他情況，則找到倒數第n個結點的前驅結點 pre；

 * 3. 移除倒數第n個結點，即 pre.next=slow.next 或 pre.next=pre.next.next ;

【代碼實現】

/**
 * 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 fast=head;		
		int step=0;
		while(step<n&&fast!=null){
			step++;
			fast=fast.next;
		}
		// the removed node is the head
		if(fast==null){
			return head.next;
		}
		ListNode pre=head;
		ListNode slow=head;
		// find the positon of the removed node , the slow node
		while(fast!=null){
			pre=slow;
			slow=slow.next;
			fast=fast.next;			
		}
		// remove the node
		pre.next=slow.next;
		return head;
    }
}