LeetCode-019 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.

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.

Analyse

一道很有趣的题，题意很简单，就是删除链表倒数第n个的数，做法可以是，先跑一遍链表数出链表长度，然后再算出顺数第几个位置再删去即可，但这种方法需要遍历两次链表，但这道题有趣的是，题目叫我们可以尝试在一次遍历解决问题。其实也很简单，可以一头一尾两个指针同时遍历链表，其中两个指针的间距为n+1，这就能保证当你尾巴那个指针到达链表尾部的时候，头那个指针刚好指的是目标的前一个位置，然后再做一个删除操作即可。

Code

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode* ans = new ListNode(-1);        ans->next = head;        ListNode* first = ans;        ListNode* second = ans;        for (int i = 0; i < n + 1; i++) {            first = first->next;        }        while (first != NULL) {            first = first->next;            second = second->next;        }        ListNode* temp = second->next;        delete(temp);        second->next = second->next->next;        return ans->next;    }};