LeetCode19-Remove-Nth-Node-From-End-of-List

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

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.

Follow up:

Could you do this in one pass?
链表操作的题，比较简单，主要主要Bug Free

public ListNode removeNthFromEnd(ListNode head, int n) {    int len=0;    ListNode cur=head;    while(cur!=null){        len++;        cur=cur.next;    }    len=len-n;    ListNode trueHead=new ListNode(0);    trueHead.next=head;    cur=trueHead;    while(len>=1){        cur=cur.next;        len--;    }    cur.next=cur.next.next;    return trueHead.next;}

上面的方法需要遍历两次链表
也可以一次遍历出

public ListNode removeNthFromEnd(ListNode head, int n) {    ListNode trueHead=new ListNode(0);    trueHead.next=head;    ListNode first=trueHead;    ListNode second=trueHead;    for(int i=0;i<n;i++) first=first.next;    while(first.next!=null){        first=first.next;        second=second.next;    }    second.next=second.next.next;    return trueHead.next;}