PAT甲级——A1133 Splitting A Linked List【25】

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 (≤) which is the total number of nodes, and a positive K (≤). The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.

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 [, 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

【题解】
新建3条链表，分别存负数，[0,k]的数以及大于k的数。然后将三条链表窜起来
当然，更省事的，就直接使用3个queue分别记录三组数据，然后输出，
两个方法差不多，容器的更方便，但链表的不用额外空间

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 struct Node
 5 {
 6     int val, next;
 7 }List[100010];
 8 int head, n, k;
 9 int main()
10 {
11     cin >> head >> n >> k;
12     while (n--)
13     {
14         int address, data, next;
15         cin >> address >> data >> next;
16         List[address].val = data;
17         List[address].next = next;
18     }
19     int head1 = 100001, head2 = 100002, head3 = 100003;//分别是负数、中间数、>k数的链表
20     int p = head, p1 = head1, p2 = head2, p3 = head3;
21     while (p != -1)
22     {
23         if (List[p].val < 0)
24         {
25             List[p1].next = p;
26             p1 = p;
27         }
28         else if (List[p].val > k)
29         {
30             List[p3].next = p;
31             p3 = p;
32         }
33         else
34         {
35             List[p2].next = p;
36             p2 = p;
37         }
38         p = List[p].next;
39     }
40     //这里的顺序千万不要反了，因为next不是地址，要先改变，再赋值
41     List[p3].next = -1;
42     List[p2].next = List[head3].next;
43     List[p1].next = List[head2].next;    
44     p = List[head1].next;
45     while (List[p].next != -1)
46     {
47         printf("%05d %d %05d\n", p, List[p].val, List[p].next);
48         p = List[p].next;
49     }
50     printf("%05d %d %d\n", p, List[p].val, List[p].next);
51     return 0;
52 }

原文地址：https://www.cnblogs.com/zzw1024/p/11488848.html