leetcode

题目：Remove Nth Node From End of List

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.

个人思路：

1、设置两个指针，一个快，一个慢，根据给定的n，让快指针先走n步，之后再同时走，直到快指针走到末节点，此时慢指针走到要删除节点的父节点

2、注意一下边界条件即可，链表为空、删除头节点等情况

代码：

#include <stddef.h>

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {};
};

class Solution
{
public:
    ListNode* removeNthFromEnd(ListNode *head, int n)
    {
        if (!head)
        {
            return NULL;
        }

        ListNode *slow = head;
        ListNode *fast = head;

        //快指针先走n步
        for (int i = 0; i < n; ++i)
        {
            fast = fast->next;
        }
        //快指针为空，表明要删除的节点为头节点
        if (!fast)
        {
            head = head->next;
            delete slow;
            slow = NULL;

            return head;
        }
        //快指针走到最后一个节点时，慢指针走到要删除节点的前一个节点
        while (fast->next)
        {
            slow = slow->next;
            fast = fast->next;
        }
        //删除节点
        ListNode *deleted = slow->next;
        slow->next = deleted->next;
        delete deleted;
        deleted = NULL;

        return head;
    }
};

int main()
{
    return 0;
}

网上的文章大部分都是这个思路