Lintcode LinkedList 174 Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

Example

Given linked list: 1->2->3->4->5->null, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5->null.

C++:

/** * Definition of ListNode * 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->next) return NULL;        ListNode *pre = head, *cur = head;        for (int i = 0; i < n ; ++i) cur = cur->next;        if (!cur) return head->next; 不用dummy        while (cur->next) {            cur = cur->next;            pre = pre->next;        }        pre->next = pre->next->next;        return head;    }};

C++ dummy版本

/** * Definition of ListNode * 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        ListNode *dummy = new ListNode(0);        dummy->next = head;        ListNode *pre = dummy;                for(int i=0;i<n;i++){            if(head==NULL)            return NULL;            head = head->next;        }                while(head!=NULL){            head = head->next;            pre = pre->next;        }                pre->next = pre->next->next;        return dummy->next;    }};


JAVA:
创建dummy一个值为0的头指针，将head连接到dummy后，如果Listnode head 为1->2->3->4->5->null,连接后的dummy为
0->1->2->3->4->5->null。注意head指针仍指向值为1的(即第二个)结点，引进dummy的原因是在1->2->null，n＝2这种情况
无需删除头结点，返回dummy.next即可解决。
head指向dummy的第二个结点，i<n-1，head循环之后指向第三个结点，pre指向第一个结点，0->1->2->3->4->5->null。
只要head.next不是null，head和pre就向后同步移动，0->1->2->3->4->5->null。然后用
pre.next = pre.next.next删除结点值为4的结点。最后返回dummy.next将值为0的结点去掉。
如果for循环中i<n,那么while循环中的条件改为head!=null,
0->1->2->3->4->5->null同步移动到0->1->2->3->4->5->null
/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } */ public class Solution {    /**     * @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 hereif (n <= 0) {        if(n <= 0) return null;        ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode pre = dummy;        for(int i = 0; i<n-1; i++){            if(head == null)            return null;            head = head.next;        }                while(head.next!=null){            head = head.next;            pre = pre.next;        }                pre.next = pre.next.next;        return dummy.next;      }}

Python
"""Definition of ListNodeclass 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        res = ListNode(0)        res.next = head        tmp = res        for i in range(0, n):            head = head.next        while head != None:            head = head.next            tmp = tmp.next        tmp.next = tmp.next.next        return res.next

类似的思路还有LinkedList166
Nth to Last Node in List，只需找到结点，无需返回链表，所以无需构造dummy。