PAT (Advanced Level) Practice 1133 Splitting A Linked List (25分) （静态链表的遍历）

1.题目

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 (≤10​5​​) 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 [−10​5​​,10​5​​], 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

2.代码

#include<iostream>
#include<vector>
using namespace std;
struct node
{
	int address;
	int data;
	int next;
}list[100001];

int main()
{
	int head, n, k;
	int temp;
	cin >> head >> n >> k;
	int headcpoy = head;
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &temp);
		scanf(" %d %d", &list[temp].data, &list[temp].next);
		list[temp].address = temp;
	}
	int count = 0; int count1 = 0, count2 = 0, count3 = 0;
	vector<node>out;
	while (1)
	{
		if (list[headcpoy].data < 0)out.push_back(list[headcpoy]);
		headcpoy = list[headcpoy].next;
		if (headcpoy == -1)break;
	}
	headcpoy = head;
	while (1)
	{
		if (list[headcpoy].data >= 0&& list[headcpoy].data<=k)out.push_back(list[headcpoy]);
		headcpoy = list[headcpoy].next;
		if (headcpoy == -1)break;
	}
	headcpoy = head;
	while (1)
	{
		if (list[headcpoy].data>k)out.push_back(list[headcpoy]);
		headcpoy = list[headcpoy].next;
		if (headcpoy == -1)break;
	}
	for (int i = 0; i < out.size() - 1; i++)
		printf("%05d %d %05d
", out[i].address, out[i].data, out[i + 1].address);
	printf("%05d %d -1
", out[out.size() - 1].address, out[out.size() - 1].data);

}