【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.

Java:

照猫画虎：

1

/**
 * 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 slow=head;
        ListNode fast=head;
        while(n>0&&fast!=null){
            fast=fast.next;
            n--;
        }
        if(n>0) return head;
        if(fast==null) return head.next;
        while(fast.next!=null){
            slow=slow.next;
            fast=fast.next;
        }
        slow.next=slow.next.next;
        return head;
    }
}

参考：征服大哥 http://blog.youkuaiyun.com/linhuanmars/article/details/19778441

public ListNode removeNthFromEnd(ListNode head, int n) {  
    if(head == null)  
        return null;  
    int i=0;  
    ListNode runner = head;  
    while(runner!=null && i<n)  
    {  
        runner = runner.next;  
        i++;  
    }  
    if(i<n)  
        return head;  
    if(runner == null)  
        return head.next;  
    ListNode walker = head;  
    while(runner.next!=null)  
    {  
        walker = walker.next;  
        runner = runner.next;  
    }  
    walker.next = walker.next.next;  
    return head;  
}  

2.

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
         // Start typing your Java solution below
        // DO NOT write main() function
        if(head==null||n<=0) return null;
        ListNode nodeS=head;//slower node
        ListNode nodeF=head;//faster node
        for(int i=1;i<=n;i++)
        {
            nodeF=nodeF.next;
        }
        if(nodeF==null) //removed node is head, so remove head
        {
            head = head.next;  
            return head;  
        }
        while(nodeF.next!=null)
        {
            nodeF=nodeF.next;
            nodeS=nodeS.next;
        }
        nodeS.next=nodeS.next.next;
        return head;
    }
}

参考伪技大哥：http://blog.youkuaiyun.com/u010500263/article/details/18429061

public class Solution {  
    public ListNode removeNthFromEnd(ListNode head, int n) {  
        ListNode nodeN = head; // slower pointer (will finally point to the target remove node-1)  
        ListNode nodeE = head; // faster pointer  
          
        if (head.next == null) return null;  
          
        // set up the distance between slower pointer and faster pointer  
        for(int i=1; i<=n; i++){  
            nodeE = nodeE.next;  
        }  
          
        if (nodeE == null) {// case of removing the head node  
            head = head.next;  
            return head;  
        }   
          
        while(nodeE.next != null){  
            nodeE = nodeE.next;  
            nodeN = nodeN.next;   
        }  
          
        nodeN.next = nodeN.next.next;  
        return head;  
    }  
}