删除链表中倒数第n个节点

174. 删除链表中倒数第n个节点

给定一个链表，删除链表中倒数第n个节点，返回链表的头节点。

 

样例

给出链表1->2->3->4->5->null和 n = 2.

删除倒数第二个节点之后，这个链表将变成1->2->3->5->null.

挑战

O(n)时间复杂度

注意事项

链表中的节点个数大于等于n

输入测试数据 (每行一个参数)

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer
     * @return: The head of linked list.
     */
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // write your code here
        if(head==null || n<1) return head;
        ListNode curr=head;
        while(curr!=null)
        {
            curr=curr.next;
            n--;
        }
        if(n==0)
        {
            return head.next;
        }
        if(n<0)
        {
            curr=head;
            while(++n!=0)
            {
                curr=curr.next;
            }
            curr.next=curr.next.next;
        }
        return head;
    }
}


######################################################

/**
 * Definition of singly-linked-list:
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @param n: An integer
     * @return: The head of linked list.
     */
    ListNode * removeNthFromEnd(ListNode * head, int n) {
        // write your code here
        if(head==NULL || n<1) return head;
        ListNode* curr=head;
        while(curr!=NULL)
        {
            curr=curr->next;
            n--;
        }
        if(n==0)
        {
            return head->next;
        }
        if(n<0)
        {
            curr=head;
            while(++n!=0)
            {
                curr=curr->next;
                
            }
            curr->next=curr->next->next;
        }
        return head;
    }
};


##############################################################
"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""

class Solution:
    """
    @param head: The first node of linked list.
    @param n: An integer
    @return: The head of linked list.
    """
    def removeNthFromEnd(self, head, n):
        # write your code here
        if head==None or n<1:
            return head
        curr=head
        res=ListNode(0)
        res.next=head
        tmp=res
        for i in range(0,n):
            curr=curr.next
            if curr==None:
                break
        while curr!=None:
            curr=curr.next
            tmp=tmp.next
        tmp.next=tmp.next.next
        return res.next