使用链表中的递归来反转k个节点的最优解

I have written the code, can anyone optimize it? If u need the full code then comment, I will do it. Help me out. I did this because I didn't get the code online for doing this in a recursive way.

我写过代码，有人可以优化吗？如果你需要完整的代码然后评论，我会这样做。帮帮我。我这样做是因为我没有在线获取代码以递归方式执行此操作。

n reversekNode(n head,n pre,n frst,int i,int m){
    if(head!=NULL){
        if(i<m-1){
            n nxt = head->next;
            head->next = pre;
            if(i==0)
                frst = head;
            return reversekNode(nxt,head,frst,i+1,m);
        }else{
            n nn = head->next;
            head->next=pre; 
            if(head->next!=NULL){
                frst->next = reversekNode(nn,head,NULL,0,m);
            }
            return head;
        }
    }
}

1 个解决方案

#1

The simplest code I can think of to recursively reverse the first k items in a list is this

我能想到的最简单的代码是递归地反转列表中的前k项是这样的

Node *reverseList(Node *a, Node *b, int k)
{
    if (b == NULL || k <= 1)
        return a;

    Node *head = reverseList(b, b->next, k-1);
    Node *temp = b->next;
    b->next = a;
    a->next = temp;
    return head;
}

It should be called like this

它应该像这样调用

if (head != NULL)
    head = reverseList(head, head->next, k);