#19 Remove Nth Node From End of List

题目链接：https://leetcode.com/problems/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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    struct ListNode* p = head;
    struct ListNode* q = head;
    while(n--)
        q = q->next;
    if(q == NULL) { //删除头结点
        free(p);
        return head->next;
    }
    while(q->next != NULL) {    //找到要删除节点的前一个节点
        p = p->next;
        q = q->next;
    }
    q = p->next;
    p->next = q->next;
    free(q);
    return head;
}

#19 Remove Nth Node From End of List

原文：http://blog.csdn.net/ice_camel/article/details/46876315