19. 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.

                    

                            死办法。。  有一点要注意，  见下面注释

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        int cnt = 0;        int flag = 0;        ListNode* p = head;        if(!head)        return NULL;        if(head->next == NULL && n == 1)        return NULL;        while(head)        {            cnt++;            head = head->next;        }        ListNode* q = p;        while(q)        {            flag++;            if(n == 1)                            //这里是判断，如果删除的是尾结点，操作会不一样，而且也没有NEXT，直接写next就错了            {                if(flag ==(cnt - n))                q->next = NULL;            }            else            if(flag == (cnt - n + 1))            {                q->val = q->next->val;                q->next = q->next->next;                break;            }            q = q->next;        }            return p;            }};

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode dum(0), *slow=&dum, *fast=head;        dum.next=head;  //get the dum connected                // move the fast pointer according to the offset        for(int i=0; i<n; ++i) fast=fast->next;        while(fast) {            fast=fast->next;            slow=slow->next;        }        //now the slow pointer will be the n+1 th node from the end (may be the dum head)        fast=slow->next;        slow->next=fast->next;        delete fast;    //remove the specified node                return dum.next;    }};