题目
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.
分析
1 典型的快慢指针的题
2 需要注意的是要先构造一个dummy来存放ListNode(0),作为头结点的头结点
为的是处理头结点被删掉的情形
3 做题时用while循环在很多时候比用for循环思路更清晰
代码
public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode firstNode = dummy; while (n > 0){ firstNode = firstNode.next; n--; } ListNode secondNode = dummy; while (firstNode.next!= null){ secondNode = secondNode.next; firstNode = firstNode.next; } secondNode.next = secondNode.next.next; return dummy.next; }