19. 删除链表的倒数第 N 个结点
链表的长度
用到哑节点
遍历到 L−n+1 个节点时，它的下一个节点就是我们需要删除的节点，只需要修改一次指针，就能完成删除操作。
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0, head);
int length = getLength(head);
ListNode cur = dummy;
for (int i = 1; i < length - n + 1; ++i) {
cur = cur.next;//知道要删除的节点的前一个
}
cur.next = cur.next.next;
ListNode ans = dummy.next;//保留最开始位置
return ans;
}
public int getLength(ListNode head) {
int length = 0;
while (head != null) {
++length;
head = head.next;
}
return length;
}
}
双指针、
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0, head);
ListNode first = head;//双指针错开一个位置.
ListNode second = dummy;//
for (int i = 0; i < n; ++i) {
first = first.next;
}
while (first != null) {//退出循环时first指向null.
first = first.next;
second = second.next;
}
second.next = second.next.next;
ListNode ans = dummy.next;
return ans;
}
}
