【leetcode】19. Remove Nth Node From End of List 删除链表的倒数第N个元素

1. 题目

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.

2. 思路

两个指针间隔距离N移动，直到前一个指针到达末尾节点。删除第一个指针后面的节点。注意好边界条件。

3. 代码

/**
 * 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 (n <= 0 || head == NULL) return head;
        ListNode* p1 = head;
        while (n > 0 && p1 != NULL) {
            p1 = p1->next;
            n--;
        }
        if (p1 == NULL) {
            return head->next;
        }
        
        ListNode* p2 = head;        
        while (p1->next != NULL) {
            p1 = p1->next;
            p2 = p2->next;
        }
        ListNode* tmp = p2->next;
        p2->next = ((tmp == NULL) ? NULL : tmp->next);
        //if (tmp != NULL) delete tmp;
        
        return head;
    }
};