02-线性结构3 Reversing Linked List (25 分)

02-线性结构3 Reversing Linked List (25 分)

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 (≤10​5​​) 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.

Sample Input:

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

Sample Output:

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

 

# include <iostream>
# include <algorithm>
# include <vector>
# include <string>
# include <map>
# include <stack>
# include <map>
using namespace std;
const int maxn = 1e5 + 3;
struct node {
	int address;
	int val;
	int next;
	node*next1;
	node(int add, int v, int next):address(add), val(v), next(next), next1(nullptr){}
	node(){}
}a[maxn];
int main() {
	int first, n, k;
	cin >> first >> n >> k;
	for (int i = 0; i < n; ++i) {
		int add, val, next;
		cin >> add >> val >> next;
		a[add].address = add;
		a[add].val = val;
		a[add].next = next;
	}
	stack<node*>s;
	node *p = new node(), *p1 = p, *p2 = p;
	while (first != -1) {
		s.push(&a[first]);
		first = a[first].next;
		if (s.size() == k) {
			while (!s.empty()) {
				node *temp = s.top();
				s.pop();
				p->next1 = new node(temp->address, temp->val, temp->next);
				p->next = p->next1->address;
				p = p->next1;
			}
		}
	}
	stack<node*>s1;
	while (!s.empty()) {
		node*temp = s.top();
		s.pop();
		s1.push(temp);
	}
	while (!s1.empty()) {
		node*temp = s1.top();
		s1.pop();
		p->next1 = new node(temp->address, temp->val, temp->next);
		p->next = p->next1->address;
		p = p->next1;
	}
	p->next1 = nullptr;
	p->next = -1;
	p2 = p2->next1;
	while (p2) {
		int s1 = p2->address;
		int s2 = p2->next;
		string ss1 = to_string(s1);
		string ss2 = to_string(s2);
		if (ss1.size() < 5) {
			for (int i = ss1.size(); i < 5; ++i)
				cout << "0";
		}
		cout << ss1 << " " << p2->val << " ";
		if (ss2.size() < 5 && ss2 != "-1") {
			for (int i = ss2.size(); i < 5; ++i)
				cout << "0";
		}
		cout << ss2 << endl;
		p2 = p2->next1;
	}
}