leetcode链表--18、remove-nth-node-from-end-of-list（从链表中删除倒数第k个结点）

题目描述

 

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.

 

解题思路：参照找倒数第k个结点那个题

1）两个指针p1 p2  p2先走k-1，然后两个指针一起走，p2走到最后，p1即为倒数第k个结点pre为倒数第k+1个结点

2）从链表中删除第k个节点

本题应注意边界条件的判断：

1）链表为空

2）要删除的正好是第一个结点的情况

/**
 * 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 == NULL)
            return NULL;
        ListNode *pPre = NULL;
        ListNode *p1 = head;
        ListNode *p2 = head;
        for(int i=0;i<n-1;i++)
        {
            p2 = p2->next;
        }
        while(p2->next != NULL)
        {
            pPre = p1;
            p1 = p1->next;
            p2 = p2->next;
        }
        if (pPre == NULL)//正好要删除的就是第一个结点
        {
            head = p1->next;
            delete p1;
        }
        else
        {
           pPre->next = p1->next;
           delete p1;
        }
 
        return head;
    }
};