PAT A1133 Splitting A Linked List (25point(s))

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Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

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 (≤$10​3$). 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 in [−$105$ ,$105$], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218
Sample Output:
33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

解答

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<unordered_map>

using namespace std;

struct node { int add, key, next; };

int main()
{
	int head, num, K, add, key, next;
	unordered_map<int, node> list;
	scanf("%d %d %d", &head, &num, &K);
	for (int i = 0; i < num; i++) {
		scanf("%d %d %d", &add, &key, &next);
		list[add] = { add, key, next };
	}
	vector<node> v, ans;

	for (int tail = head; tail != -1; tail = list[tail].next)
		v.push_back(list[tail]);

	for (int i = 0; i < v.size(); i++) 
		if (v[i].key < 0) ans.push_back(v[i]);
	for (int i = 0; i < v.size(); i++) 
		if (v[i].key >= 0 && v[i].key <= K) ans.push_back(v[i]);
	for (int i = 0; i < v.size(); i++)
		if (v[i].key > K) ans.push_back(v[i]);

	for (int i = 0; i < v.size() - 1; i++)
		printf("%05d %d %05d\n", ans[i].add, ans[i].key, ans[i + 1].add);
	printf("%05d %d -1\n", ans.back().add, ans.back().key);
	return 0;
}