Leetcode[19]-Remove Nth Node From End of List

Link: https://leetcode.com/problems/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.

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思路：先将链表翻转，然后遍历，找到第n-1个节点，然后删除第n个节点，最后再次翻转链表即可。

笔记：在链表翻转后，我采用的是在翻转后的链表的头部放一个无关的节点，然后往后面找，变量从1到n，当变量为n的时候，此时节点还处于n-1节点上，接着我们就让该节点的下个节点指向它的下下个节点，这样第k个节点就删除了。

Code(c++):

/** * 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) {        reverseList(head);        ListNode *newList = new ListNode(-1);        newList->next = head;        int count = 1;        ListNode *pre = newList;        //find the prior element            //when we get the Nth Node,delete this node from list            if(count==n && pre->next!=NULL){                pre->next = pre->next->next;                break;            }else{                pre = pre->next;                count++;            }        }        head = newList->next;        //when head is not null,reverse it        if(head!=NULL)            reverseList(head);        return head;    }    void reverseList(ListNode* &head){        if(head==NULL || head->next == NULL)return;        ListNode *newList = new ListNode(-1);        ListNode *pre = head;        ListNode *temp;        while(pre!=NULL){            temp = pre->next;            pre->next = newList->next;            newList->next = pre;            pre = temp;        }        newList = newList->next;        head = newList;    }};