Remove Nth Node From End of List--LeetCode

1.题目

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.

2.题意

只遍历一次链表，删除链表中倒数第N个元素

3.分析

只允许遍历一次
所以我们不能用一次完整的遍历来统计链表长度
而是遍历到对应位置就应该移除了
可以使用快慢指针
cur先走n步后，prev、cur再一起走
当cur->next为空删掉prev->next即可
时间复杂度为O(链表的长度)，空间复杂度为O(1)
两个需要注意的点
1)test中似乎没有给出i < n的测试数据
2)但是注意cur == nullptr时删去首节点

4.代码

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        if(head == nullptr)            return nullptr;        int i = 0;        ListNode *cur = head;        while(i < n && cur != nullptr)        {            cur = cur->next;            ++i;        }        // if(i < n)        //     return head;        if(cur == nullptr)            return head->next;        ListNode *prev = head;        while(cur->next != nullptr)        {            cur = cur->next;            prev = prev->next;        }        ListNode *tmp = prev->next;        prev->next = prev->next->next;        delete tmp;        return head;    }};