19. Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

t_jianyan

最新推荐文章于 2024-01-12 15:01:11 发布

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分类专栏： leetcode代码笔记

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本文介绍了一种从链表中删除倒数第N个节点的方法，并提供了两种不同的实现方式。一种是通过遍历整个链表来确定位置并进行删除，另一种则采用双指针技巧简化操作。

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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.

死办法。。有一点要注意，见下面注释

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int cnt = 0;
        int flag = 0;
        ListNode* p = head;
        if(!head)
        return NULL;
        if(head->next == NULL && n == 1)
        return NULL;
        while(head)
        {
            cnt++;
            head = head->next;
        }
        ListNode* q = p;
        while(q)
        {
            flag++;
            if(n == 1)                            //这里是判断，如果删除的是尾结点，操作会不一样，而且也没有NEXT，直接写next就错了
            {
                if(flag ==(cnt - n))
                q->next = NULL;
            }
            else
            if(flag == (cnt - n + 1))
            {
                q->val = q->next->val;
                q->next = q->next->next;
                break;
            }
            q = q->next;
        }
            return p;
        
    }
};

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode dum(0), *slow=&dum, *fast=head;
        dum.next=head;  //get the dum connected
        
        // move the fast pointer according to the offset
        for(int i=0; i<n; ++i) fast=fast->next;
        while(fast) {
            fast=fast->next;
            slow=slow->next;
        }
        //now the slow pointer will be the n+1 th node from the end (may be the dum head)
        fast=slow->next;
        slow->next=fast->next;
        delete fast;    //remove the specified node
        
        return dum.next;
    }
};