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.
题目含义:删除列表中从末尾数第n个节点
1 public ListNode removeNthFromEnd(ListNode head, int n) { 2 ListNode start = new ListNode(0); 3 ListNode slow = start, fast = start; 4 start.next = head; 5 6 for (int i = 0; i <= n ; i++) { 7 fast = fast.next; 8 } 9 while (fast != null) { 10 slow = slow.next; 11 fast = fast.next; 12 } 13 slow.next = slow.next.next; 14 return start.next; 15 }