LeetCode--Remove Nth Node From End of List

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题目描述：

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

note：

Given n will always be valid.

样例：

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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.

题意：

给出一个链表，移除并返回删除后的链表

思路：

直接两个指针 p、q，p 先遍历前面 n 个节点，然后 p、q 一起遍历，直到 p 遍历到最后一个节点，删除当前的 q 节点

代码：

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/**
 * 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) {
        if(n == 0) return head;
        if(head == NULL) return head;
        ListNode *pl = head, *pr = head;
        for(int i = 1; i <= n; ++i) pr = pr->next;
        if(pr == NULL) {
            head = head->next;
            return head;
        }
        while(pr->next != NULL){
            pr = pr->next;
            pl = pl->next;
        }
        pl->next = pl->next->next;
        return head;
    }
};

Runtime：8ms Memory：9.5MB