019 删除链表的倒数第N个节点

LeetCode 第十九题 删除链表的倒数第N个节点

给定一个链表，删除链表的倒数第 n 个节点并返回头结点。
例如，
给定一个链表: 1->2->3->4->5, 并且 n = 2.

当删除了倒数第二个节点后链表变成了 1->2->3->5.

说明:
给的 n 始终是有效的。
尝试一次遍历实现。

在这里我们参考了网址的解题算法。

方法一

需要注意的是，删除倒数第N个节点，其实就是删除第(L-n+1)个节点，这里L表示整个链表的长度，我们可以首先遍历一遍链表，计算出链表的长度，接着根据L-n+1找到该节点，做删除操作。

Java

    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        int length = 0;
        ListNode first = head;
        while (first != null) {
            length++;
            first = first.next;
        }
        length -= n;
        first = dummy;/// 确保了可以删除头节点的操作
        while (length > 0) {
            length--;
            first = first.next;
        }
        first.next = first.next.next;
        return dummy.next;
    }

Python

class Solution(object):
    def removeNthFromEnd(self, head, n):
        dummy = ListNode(0)
        dummy.next = head
        length = 0
        first = head
        while (first):
            length = length + 1
            first = first.next
        length = length - n
        first = dummy
        while (length > 0):
            length = length - 1
            first = first.next
        will_delete = first.next
        first.next = will_delete.next
        del (will_delete)
        return dummy.next

C++

ListNode* removeNthFromEnd(ListNode* head, int n)
    {
        ListNode dummy = ListNode(0);
        dummy.next = head;
        int length = 0;
        ListNode* first = head;
        while(first!=NULL)
        {
            length++;
            first = first->next;
        }
        length -= n;
        first = &dummy;
        while(length>0)
        {
            length--;
            first = first->next;
        }
        ListNode *will_delete = first->next;
        first->next = will_delete->next;
        delete(will_delete);
        return dummy.next;
    }

方法二

我们可以使用两个指针first和second，两个指针在初始状态时，都指向头节点，接着移动first指针，使其向后移动n个节点，这样first和second指针之间就间隔了n个节点，接着同时移动first和second指针（类似于平移的操作），直到first指针指向了尾节点，此时，second指针刚好指向了倒数第N个节点，将其删除。该算法十分巧妙，很喜欢。

Java

    public ListNode removeNthFromEnd_2(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = dummy;
        for (int i = 0; i <= n; i++) {
            first = first.next;
        }
        while (first != null) {
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return dummy.next;
    }

Python

    def removeNthFromEnd_2(self, head, n):
        dummy = ListNode(0)
        dummy.next = head
        first = dummy
        second = dummy
        for i in range(n + 1):
            first = first.next
        while (first):
            first = first.next
            second = second.next
        will_delete = second.next
        second.next = will_delete.next
        del (will_delete)
        return dummy.next

C++

 ListNode* removeNthFromEnd_2(ListNode* head, int n)
    {
        ListNode dummy(0);
        dummy.next=head;
        ListNode* first=&dummy;
        ListNode* second=&dummy;
        for(int i=0;i<=n;i++)
        {
            first=first->next;
        }
        while(first!=NULL)
        {
            first=first->next;
            second=second->next;
        }
        ListNode* will_delete = second->next;
        second->next=will_delete->next;
        delete(will_delete);
        return dummy.next;
    }