LeetCode | 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. Note: Given n will always be valid. Try to do this
in one pass.

这道题思路倒是想到了，
刚刚一直卡在两个指针怎么互传值的

= =||
这个删元素的办法
t1->next=t1->next->next;

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode root(-1);        root.next=head;        ListNode* t1=&root;        ListNode* t2=t1;        int count=0;        for(;t2->next!=NULL;){            if(count<n){                t2=t2->next;                count++;            }            else{                t1=t1->next;                t2=t2->next;            }        }        //删除某一元素神器        t1->next=t1->next->next;        // if(count==1 && head->next==NULL) return NULL;        // if(t1==&root && count==0) t1->next=head->next;        // else t1->next=t2==NULL?NULL:t2->next;        return root.next;    }};