Remove Nth Node From End of List (Easy)

Description

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.

Analysis

一次循环删除链表中倒数第n个元素。维护两个指针。之间的距离为n，
则当一个指针到达尾部时，另一个指针指向的则为要删除的节点

My Solution

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//C++
/**
 * 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 *p = head,*q = head;
        int dist = 0;
        while(p->next!=NULL){
            p = p->next;
            dist++;
            if(dist>n)
                q = q->next;
        }
        if(dist<n){
            p = head;
            head = head->next;
            free(p);
        }
        else{
            p = q->next->next;
            free(q->next);
            q->next = p;
        }
        
        return head;
    }
};