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

 #include <iostream>
 #include <vector>
 using namespace std;
 struct Node
 {
     int val, next;
 }List[];
 int head, n, k;
 int main()
 {
     cin >> head >> n >> k;
     while (n--)
     {
         int address, data, next;
         cin >> address >> data >> next;
         List[address].val = data;
         List[address].next = next;
     }
     int head1 = , head2 = , head3 = ;//分别是负数、中间数、>k数的链表
     int p = head, p1 = head1, p2 = head2, p3 = head3;
     while (p != -)
     {
         if (List[p].val < )
         {
             List[p1].next = p;
             p1 = p;
         }
         else if (List[p].val > k)
         {
             List[p3].next = p;
             p3 = p;
         }
         else
         {
             List[p2].next = p;
             p2 = p;
         }
         p = List[p].next;
     }
     //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
     List[p3].next = -;
     List[p2].next = List[head3].next;
     List[p1].next = List[head2].next;
     p = List[head1].next;
     while (List[p].next != -)
     {
         printf("%05d %d %05d\n", p, List[p].val, List[p].next);
         p = List[p].next;
     }
     printf("%05d %d %d\n", p, List[p].val, List[p].next);
     return ;
 }