[LeetCode]19. 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;
}