PAT 乙级 1015 反转链表

题目描述：

给定一个常数K以及一个单链表L，请编写程序将L中每K个结点反转。例如：给定L为1→2→3→4→5→6，K为3，则输出应该为

3→2→1→6→5→4；如果K为4，则输出应该为4→3→2→1→5→6，即最后不到K个元素不反转。

输入描述 

每个输入包含1个测试用例。每个测试用例第1行给出第1个结点的地址、结点总个数正整数N(<= 105)、以及正整数K(<=N)，即要求反转的

子链结点的个数。结点的地址是5位非负整数，NULL地址用-1表示。
接下来有N行，每行格式为：
Address Data Next
其中Address是结点地址，Data是该结点保存的整数数据，Next是下一结点的地址。

输出描述 

对每个测试用例，顺序输出反转后的链表，其上每个结点占一行，格式与输入相同。

我的代码结果正确，但超时了。。

#include <iostream>
#include <stack>
#include <vector>
#include <string>
using namespace std;

class linknode {
public:
	string previous;
	int data;
	string next;
	linknode(string previous, int data, string next)
	{
		this->data = data;
		this->next = next;
		this->previous = previous;
	}
};



int main(void) {
	vector<linknode> nodes_vector;
	vector<linknode> nodes_temp_vector;
	string next_node;
	int n, k;
	string previous, next;
	int data;
	int index=0;

	cin>> next_node>>n>>k;
	for (int i = 0; i < n; ++i) {
		cin>> previous >>data>>next;
		linknode node(previous, data, next);
		nodes_vector.push_back(node);
	}
	while(next_node!="-1")
	{
		for (int i = 0; i < n; ++i)
		{
			linknode node = nodes_vector[i];
			if (node.previous == next_node)
			{
				nodes_temp_vector.push_back(node);
				next_node = node.next;
				break;
			}
		}
	}
	
	nodes_vector.clear();
	stack<linknode> nodes_stack;
	for (int i = 0; i < (n / k); i++)
	{
		for (int j = 0; j < k; j++)
		{
			nodes_stack.push(nodes_temp_vector[index]);
			index++;
		}
		for (int j = 0; j < k; j++)
		{
			nodes_vector.push_back(nodes_stack.top());
			nodes_stack.pop();
		}
	}
	for (int i = index; i < n; i++)
	{
		nodes_vector.push_back(nodes_temp_vector[i]);
	}
	for (int i = 0; i < n; i++)
	{
		cout << nodes_vector[i].previous <<" "<< nodes_vector[i].data <<" "
			<<nodes_vector[i].next<< endl;		
	}


	return 0;
}

 贴一份牛客网上面的回答

Cataliar:

链接：https://www.nowcoder.com/questionTerminal/5f44c42fd21a42b8aeb889ab83b17ad0
来源：牛客网

#include <stdio.h>
#include <memory.h>
 
#define LIMIT 100002
 
struct linknode {
    int data;
    int next;
};
 
int reversek(struct linknode *, int, int);
 
int main(void) {
    struct linknode list[LIMIT];
    int faddr, n, k;
    int addr, data, next;
    int i, j, count = 0;
 
    scanf("%d %d %d", &faddr, &n, &k);
    memset(list, 0, LIMIT * sizeof(struct linknode));
    list[LIMIT-1].next = faddr;  //head node
    for(i = 0; i < n; ++i) {
        scanf("%d %d %d", &addr, &data, &next);
        list[addr].data = data;
        list[addr].next = next;
    }
    for(i = list[LIMIT-1].next; i != -1; i = list[i].next)
        ++count; //count valid nodes
    for(j = count / k, i = LIMIT - 1; j > 0 && i != -1; --j)
        i = reversek(list, i, k); //reverse every k nodes
    for(i = list[LIMIT-1].next; list[i].next != -1; i = list[i].next)
        printf("%05d %d %05d\n", i, list[i].data, list[i].next);
    printf("%05d %d %d\n", i, list[i].data, list[i].next);
    return 0;
}
 
int reversek(struct linknode *list, int head, int k) {
    int i, j, p;
    int nexthead;
    for(j = 0, p = list[head].next; j < k && p != -1; ++j) //find tail
        p = list[p].next;
    nexthead = list[head].next;
    list[head].next = p;
    for(j = 0, i = nexthead; j < k && i != -1; ++j) {
        p = list[i].next;
        list[i].next = list[head].next; //insert at the head
        list[head].next = i;
        i = p;
    }
    return nexthead;
}