leetcode 19 -- Remove Nth Node From End of List

Remove Nth Node From End of List

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.

题意：
单向链表，删除指定倒数第n个节点。

思路：
先计算链表节点的个数，然后通过计算找到将要删除的节点删除即可。注意的就是链表为空的情况和删除头节点的情况。

/**
 * 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) {
        //链表为空的情况
        if(head == NULL){
            return head;
        }
        int cnt = 0;
        ListNode *p = head;
        ListNode *q = head;
        //计数
        while(p != NULL){
            ++cnt;
            p = p->next;
        }
        //找要删除的节点的前一个节点，方便删除指定节点
        for(int i = 0; i < cnt-n-1; ++i){
            q = q->next;
        }
        //如果删除的是首节点的情况
        if(n == cnt){
            return head->next;
        }
        ListNode *temp;
        temp = q->next;
        q->next = temp->next;
        free(temp);
        return head;
    }
};