PTA线性结构3 Reversing Linked List【直接贴答案将不予采纳，谢谢合作】

Question

#include <iostream>
using namespace std;

struct Node
{
    int adrs;
    int data;
    int next;
    struct Node* link;
};


Node* reverselink(struct Node* node, int K)
{
    struct Node* initnode = node;
    struct Node* newnode = node -> link;
    struct Node* oldnode = newnode -> link;
    struct Node* temp;

    int count = 1;

    while(count < K)
    {
        temp = oldnode -> link;
        oldnode -> link = newnode;
        newnode = oldnode;
        oldnode = temp;
        count++;
    }

    initnode -> link -> link = oldnode;

    return newnode;
}

void print(struct Node* initNode)
{
    while(initNode != NULL)
    {
        if(initNode -> link != NULL)
        {
            printf("%05d %d %05d\n", initNode -> adrs, initNode -> data, initNode -> link -> adrs);

        }
        else
        {
            printf("%05d %d %d\n", initNode -> adrs, initNode -> data, -1);
        }

        initNode  = initNode -> link;
    }
}

int main()
{
    int start = 0;
    int N = 0;
    int K = 0;
    scanf("%d %d %d",&start, &N, &K);
    int adrs = -1;
    int cnt = 0;

    int data[100000] = {0};
    int next[100000] = {0};
    while(cnt < N)
    {
        scanf("%d", &adrs);
        scanf("%d", &data[adrs]);
        scanf("%d", &next[adrs]);
        cnt++;
    }
    //解决多余结点
    int actualNum = 0;
    struct Node* initNode = new Node;
    struct Node* prevNode = initNode;
    struct Node* curNode = NULL;

    for(int i = start; i <= 100000 && i != -1; )
    {
        curNode = new Node;
        curNode -> data = data[i];
        curNode -> adrs = i;
        curNode -> next = next[i];
        curNode -> link = NULL;//////
        prevNode -> link = curNode;
        prevNode = curNode;
        i = next[i];
        actualNum++;
    }
    cnt = 0;
    N = actualNum;

    curNode -> link = NULL;

    if(K == 1)//直接输出
    {
        initNode = initNode -> link;
    }
    else if(K == N || (N / K == 1 && N % K != 0))//全反转  有尾巴不反转
    {
        initNode = reverselink(initNode, K);
    }
    else if(N % K == 0 && N / K != 1)
    {
        int count = 1;
        int flag = 0;
        struct Node* markNode = initNode;
        struct Node* firstNode = initNode;
        while(cnt < N / K)
        {
            initNode = reverselink(initNode, K);
            markNode -> link = initNode;
            cnt++;
            while(count < K && cnt < N / K)
            {
                initNode = initNode -> link;
                count++;
            }
            markNode = initNode;
            count = 1;
        }
        initNode = firstNode -> link;
    }
    print(initNode);
}

请勿贴他人的代码，贴他人的代码及网址回答将不会采纳。

Answer 1

题目已经自己解答出来，只需要把

else if(N % K == 0 && N / K != 1)

的N% K==0删除即可
应该说是N%K有余数也需要按照反转多次的办法来