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[100010];
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 = 100001, head2 = 100002, head3 = 100003;//分别是负数、中间数、>k数的链表
    int p = head, p1 = head1, p2 = head2, p3 = head3;
    while (p != -1)
    {
        if (List[p].val < 0)
        {
            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 = -1;
    List[p2].next = List[head3].next;
    List[p1].next = List[head2].next;    
    p = List[head1].next;
    while (List[p].next != -1)
    {
        printf("%05d %d %05d
", p, List[p].val, List[p].next);
        p = List[p].next;
    }
    printf("%05d %d %d
", p, List[p].val, List[p].next);
    return 0;
}