LeetCode #19 Remove Nth Node From End of List

Description

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.

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Analysis

题目难度为：Medium
题目其实就是找到单向链表中的倒数第n位，并将其删除。我们需要两个指针，先把指针1移动到正数第n位，在同时移动两个指针，指针1指向末尾的时候，指针2就指向了倒数第n位。想要删除第n位，我们需要指针指向第n-1位，所有开始时候，可以让指针2指向第0位，也就是首位的前一位，再进行刚刚的操作，就可以找到第n-1位，然后将第n位删除，返回链表

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Code（c++)

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode* p, *work;        ListNode tmp(0);        tmp.next = head;        p = head;        work = &tmp;        while (n--) {            p = p->next;        }        while (p != NULL) {            p = p->next;            work = work->next;        }        p = work->next;        if (p == head) return head->next;        work->next = p->next;        p->next = NULL;        return head;    }};