1074. Reversing Linked List (25)

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 10⁵) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std;
struct Node {
	int adddress;
	int value;
	int next;
};
vector<Node> list;
int val[100001] = { 0 }, nex[100001] = { 0 };
int main(void) {
	int first_addr, n, k;
	cin >> first_addr >> n >> k;
	if (first_addr == -1) //首地址为-1.直接输出  
	{
		printf("-1\n");
		return 0;
	}
	for (int i = 0; i < n; i++) {
		int temp_addr;
		cin >> temp_addr;
		cin >> val[temp_addr] >> nex[temp_addr];
	}
	int addr = first_addr;
	
	while (1) {
		Node node_temp;
		node_temp.adddress = addr;
		node_temp.value = val[addr];
		node_temp.next = nex[addr];
		addr = nex[addr];
		list.push_back(node_temp);
		if (addr == -1)
			break;
	}
	while (k > list.size()) {
		reverse(list.begin(), list.end());
		k -= list.size();
	}
	/*reverse(list.begin(), list.begin() + k);*/
	int loop = list.size() / k, i;
	for (i = 0; i<loop; ++i)
	    reverse(list.begin() + i*k, list.begin() + (i + 1)*k);
	for (int i = 0; i < list.size()-1; i++) {
		list[i].next = list[i + 1].adddress;
	}
	list[list.size() - 1].next = -1;
	for (int i = 0; i < list.size(); i++) {
		if (list[i].next != -1)
			printf("%05d %d %05d\n", list[i].adddress, list[i].value, list[i].next);
		else
			printf("%05d %d %d\n", list[i].adddress, list[i].value, list[i].next);
	}
	return 0;
}