删除链表的倒数第N个元素。
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:
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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.
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Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
对于链表的删除操作我们通常需要使用前面一个节点，将前一个节点的next指针指向要删除节点的next节点。
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dummy.next = head;
preNode = dummy;
preNode.next = preNode.next.next;
return dummy.next;
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public ListNode (ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode fast = dummy;
ListNode slow = dummy;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return dummy.next;
}
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