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.

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


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;
    }
};

你可能感兴趣的:(leetcode)