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  • leetcode 题解 || Remove Nth Node From End of List 问题

    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.

    删除单链表的倒数第n个节点


    thinking:

    (1)这里的 head 是头指针。指向第一个结点!!

    !别搞混了。

    (2)为了避免反复计数,採用双指针,先让第一个指针走n-1步,再一起走,这样,等前面指针走到最后一个非空结点时。后面一个指针正好指向待删除结点的前驱!!!

    (3)延伸:


    头结点不是必须的,一般不用。经常使用的是用一个头指针head指向第一个元素结点!!

    !。!这道题就是!!!!

    /**
     * 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) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            if (head == NULL)
                return NULL;
    
            ListNode *pPre = NULL;
            ListNode *p = head;
            ListNode *q = head;
            for(int i = 0; i < n - 1; i++)
                q = q->next;
    
            while(q->next)
            {
                pPre = p;
                p = p->next;
                q = q->next;
            }
    
            if (pPre == NULL)
            {
                head = p->next;
                delete p;
            }
            else
            {
                pPre->next = pPre->next->next;
                delete p;
            }
    
            return head;
        }
    };
    




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  • 原文地址:https://www.cnblogs.com/brucemengbm/p/6802880.html
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