LeetCode#19. Remove Nth Node From End of List

• 题目：删除链表倒数第n个元素
• 难度：Medium
• 思路：这道题与查找倒数第n个结点类似，因为我们需要删除倒数第n个结点，所以我们需要找到倒数第n个结点的前一个节点（定义个fast和一个slow两个指针，fast先走n步）
• 代码：
  不给链表加上一个头结点

public class Solution {     public ListNode removeNthFromEnd(ListNode head, int n) {        if(head == null || n == 0){            return null;        }        if(head.next == null){            if(n == 1){                return null;            }else{                return head;            }        }        ListNode slow = head;        ListNode fast = head;        //fast指针先走n步        while(fast != null && n > 0){            fast = fast.next;            n--;        }        //如果n大于链表节点个数，则直接返回链表        if(n > 0){            return head;        }        //这个if判断是删除结点正好是头结点        if(fast == null && n == 0){            slow = new ListNode(head.next.val);            slow.next = head.next.next;            return slow;        }        //两个节点同时从当前位置走，直到fast指针走到最后一个节点，这时slow指向待删除节点的前一个        while(fast != null && fast.next != null){            fast = fast.next;            slow = slow.next;        }        if(slow.next != null){            ListNode delete = slow.next;            slow.next = delete.next;        }        return head;    }}

给链表加上头结点

public class Solution {     public ListNode removeNthFromEnd(ListNode head, int n) {        if(head == null || n == 0){            return null;        }        ListNode start = new ListNode(0);        ListNode slow = start;        ListNode fast = start;        slow.next = head;        //fast指针先走n步        while(fast != null && n > 0){            fast = fast.next;            n--;        }        //如果n大于链表节点个数，则直接返回链表        if(n > 0){            return head;        }        //找到待删除节点的上一个节点        while(fast != null && fast.next != null){            fast = fast.next;            slow = slow.next;        }        slow.next = slow.next.next;        return start.next;    }}