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;

    }

};