leetcode--Remove Nth Node From End of List

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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.

算法：双指针

/** * 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 p=head;        for(int i=1;i<n;i++){            p=p.next;        }        ListNode s,q;        s=null;        q=head;        while(p.next!=null){            s=q;            q=q.next;            p=p.next;        }        if(s==null)            return q.next;        s.next=q.next;        q.next=null;        q=null;        return head;    }}

c++：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *removeNthFromEnd(ListNode *head, int n) {        if(!head)            return NULL;        ListNode *p,*q,*s;        p=head;        // head已遍历,k从1开始        int k=1;        while(k<n){            p=p->next;            k++;        }        q=head;        s=NULL;        while(p->next){            s=q;            q=q->next;            p=p->next;        }        if(s){            s->next=q->next;            delete q;            return head;        }        // 一个节点,而且n==1        return q->next;    }};

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