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19. Remove Nth Node From End of

19. Remove Nth Node From End of

作者: 白菜炖豆腐 | 来源:发表于2016-05-20 14:59 被阅读0次

    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.

    分析

    题目给了一个链表的一维数组,和一个整数n,要求删除从末尾开始数第n个节点,然后返回链表的头。做法是,用两个指针,一个fast指针,一个solw指针,让slow指针永远慢fast指针n个节点。fast走完后,再用slow指针来删除需要删除的节点

    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) {
            ListNode* h = new ListNode(0);
            h->next = head;
            ListNode* slow = h;
            ListNode* fast = h;
            while(fast->next!=NULL){
                if(n<=0){
                    slow = slow->next;
                }
                fast= fast->next;
                n--;
            }
            if(slow->next!=NULL){
                slow->next = slow->next->next;
            }
            return h->next;
        }
    };
    

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