PAT甲级 PAT1133 Splitting A Linked List (25分)

PAT1133 Splitting A Linked List (25分)

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

分析

排序题，用vector放入建立的链表，排序后进行输出。

思路

1.数据结构：一个结构体，放入address，data，next和index，其中index用于按下标排序。
2.用一个map来建立adr和node的对应关系。
3.通过一个while循环建立链表
4.写出cmp方法并排序后输出。按题意，其中cmp的排序方法为：

• 负的在非负数前面。
• 比k小的排在前面。
• 如果同号的话，则按index排序。

代码

#include<iostream>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
#include<math.h>
#include<map>
#include<stack>
#include<string>
#include<limits.h>
using namespace std;

int start, n, k;

typedef struct node {
	int adr;
	int data;
	int next;
	int idx;
}node;

vector<node> list;
map<int, node> m;

//排序1 负的比正的大 排序2 比k小的比较大 排序3 若同号 则按idx排
bool cmp(node a, node b)
{
	if ((a.data < 0 && b.data < 0)) return a.idx < b.idx;
	else if (a.data < 0 && b.data >= 0) return 1;
	else if (a.data >= 0 && b.data < 0) return 0;
	else if (a.data >= 0 && b.data >= 0) {
		if (a.data <= k && b.data > k) return 1;
		else if (a.data > k&&b.data <= k) return 0;
		else return a.idx < b.idx;
	}
}

int main()
{
	int adr, data, next;
	int root, num = 0;
	cin >> start >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> adr >> data >> next;
		node n = { adr,data,next };
		m[adr] = n;
	}
	while (start != -1) {
		m[start].idx = num++;
		list.push_back(m[start]);
		start = m[start].next;
	}
	sort(list.begin(), list.end(), cmp);
	for (int i = 0; i < list.size(); i++) {
		if (i == 0) printf("%05d %d", list[i].adr, list[i].data);
		else printf(" %05d\n%05d %d", list[i].adr, list[i].adr, list[i].data);
		if (i == list.size() - 1) printf(" -1\n");
	}
	system("pause");
	return 0;
}

改进

想法：用一个优先队列存储，插入的时候即实现排序，这种方法需要在结构体里面重载运算符，不赘述。