【LeetCode】之Remove Nth Node From End of List

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.

此题目的在于将一个链表的倒数第n个节点删除。

思路：
1.初始化两个节点p，q，p->next和q->next都指向head；
2.先让p节点往后走n步，然后让p，q节点同时往后走，直到p->next是NULL；
3.此时q->next就是链表的倒数第n个节点，删除这个节点，将链表连接起来;
4返回这个新链表就是最后的结果。

/** * 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) {        ListNode dummy(-1);        ListNode *p = &dummy;        ListNode *q = &dummy;        dummy.next = head;//p->next和q->next都指向head        while(n--){//p节点往后走n步            p = p->next;        }        while(p->next){//p，q节点同时往后走，直到p->next是NULL            p = p->next;            q = q->next;        }        ListNode *temp = q->next;//q->next就是链表的倒数第n个节点        q->next = q->next->next;//链表连接起来        delete temp;//删除这个节点        return dummy.next;    }};