Remove Nth Node From End of List

Question: 

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.

Solution:

两个指针，距离相差n。下面的代码考虑了n异常的情况。代码在leetcode上做了验证。

/**
 * 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 (!head)
            return head;
        ListNode *node = new ListNode(0);
        node->next = head;
        ListNode *p1, *p2;
        p1 = p2 = node;
        for (int i = 0; i <n; i++) {
            p2 = p2->next;
            if (!p2)
                return head;
        }
        while (p2->next) {
            p2 = p2->next;
            p1 = p1->next;
        }
        p2 = p1->next;
        p1->next = p2->next;
        if (p2)
            delete p2;
        p1 = node->next;
        delete node;
        return p1;
    }
};