leetcode 19. Remove Nth Node From End of List（链表）

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.

简单的指针操作。

/**
 * 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) {
        ListNode *pn=head;
        int sum=1;
        while(pn->next){
            pn=pn->next;
            sum++;
        }
        int tn=sum-n+1;
        if(tn==1){
            return head->next;
        }
        pn=head;
        int i=1;
        while(i!=tn-1){
            pn=pn->next;
            i++;
        }
        pn->next = pn->next->next;
        return head;
        
    }
};