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文章目录
  1. 1. Remove Nth Node From End of List
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题

Remove Nth Node From End of List

题目

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.

思路

主要是要将指针指到链表从末尾向前数的第n+1个结点P，知道P结点，就能通过P->next=p->next->next。不过要注意一种特定情形，删除的节点刚好是首结点。可以通过定义头结点来解决。通过两个指针来解决，一个指针先行n+1步，然后再两个指针同时移步。

解题

c++版

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode dummy(-1);
        dummy.next=head;
        ListNode *p,*q;
        p=&dummy;
        q=&dummy;
        while(n--)
            p=p->next;
        p=p->next;
        while(p and q){
            p=p->next;
            q=q->next;
        }
        q->next=q->next->next;
        return dummy.next;
    }
};

python版

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} head
    # @param {integer} n
    # @return {ListNode}
    def removeNthFromEnd(self, head, n):
        dummy=ListNode(0)
        dummy.next=head
        p=dummy
        q=dummy
        for i in range(n+1):
            p=p.next
        while(p and q):
            p=p.next
            q=q.next
        q.next=q.next.next
        return dummy.next
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文章目录
  1. 1. Remove Nth Node From End of List
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题