LeetCode OJ刷题历程——Remove Nth Node From End of List

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Given a linked list, remove the nth node from the end of list and return its head.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.

Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};

以上是题目要求，下面贴出代码：

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode *p,*q,*r;int count = 0;r = p = q = head;while(p->next){p = p->next;if(count>=n-1)q = q->next;if(count>=n)r = r->next;count++;}if(count == 0)    head = NULL;else if(r == q)    head = head->next;r->next = q->next;q->next = NULL;return head;    }};

这里只使用了一次遍历，p指针开始遍历，q指针相对于p延迟n-1个时间开始遍历，用以得到倒数第n个结点位置，而r指针用以得到倒数第n+1个结点位置，用以删除操作。

这个题目同样需要注意特殊情况，如链表只有一个结点的情况，应该返回NULL

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0,1].

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