[C++]LeetCode 19: Remove Nth Node From End of List（删除链表中倒数第n个节点）

Problem：
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.
分析：
也是出现过很多次的基本算法题。
考察点是两指针的联合使用，先让指针1走n步，再一起走，直到指针1的next为NULL，此时指针2的next为要删除的节点。两指针联合使用的经典问题还有判断链表是否有环。
题目中说了Given n will always be valid.，但码代码的时候为保持完整性，还是考虑了invalid情况。

AC Code(C++):

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //207 / 207 test cases passed.
    //Runtime: 8 ms
    
    //虽然题设说了Given n will always be valid，但为保持完整性还是考虑了invalid的情况
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == NULL || n < 1){//无效输入
            return head;
        }
        
        ListNode *pNode1 = head;
        //pNode1先走n步
        for (int step = 0; step < n; ++step) {
            if (pNode1 == NULL) {//n无效
                return head;
            }
            pNode1 = pNode1->next;
        }
        if (pNode1 == NULL) {//要删除的是第一个节点
            ListNode *pTemp = head->next;
            delete head;
            return pTemp;
        }
    
        //两指针一起走，当pNode1走到pNode1->next == NULL时，pNode2走到倒数第n+1
        ListNode *pNode2 = head;
        while (pNode1->next != NULL) {
            pNode1 = pNode1->next;
            pNode2 = pNode2->next;
        }
        
        //pNode2的next为要删除节点
        ListNode *pTemp = pNode2->next;
        pNode2->next = pNode2->next->next;
        delete pTemp;
        return head;
    }
};