[LeetCode] Remove Nth Node From End of List

Remove Nth Node From End of List

写完最长前缀，看了一眼下一个easy，发现是个几行代码的题。so，写了吧。
题目：

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.

• 解法：先用一个指针p指向头节点，然后把头节点head移动n次，然后再两个指针一起移动直到head移动到结尾，最后删除p的next就是倒数第n个节点了(p到head的距离为n).

/** * 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* pkResult = new ListNode(0);        pkResult->next = head;//新建一个节点并把next指向head，方便后面返回.        ListNode* p = pkResult;        for (; n--; head = head->next);//把head节点移动n次.        for (; head; head = head->next, p = p->next);//没移动到尾，就继续往下移动，但这些p也要跟着移动.        p->next = p->next->next;//把p的next删除.        return pkResult->next;    }};

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