1133. Splitting A Linked List (25)

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本文介绍了一个算法问题，即如何对给定的单链表进行重组，使得所有负数出现在非负数之前，并且所有在[0,K]范围内的值出现在大于K的值之前，同时保持各分类内部顺序不变。

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 (<= 10⁵) which is the total number of nodes, and a positive K (<=1000). 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 [-10⁵, 10⁵], 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

#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;

struct node
{
    int address,data,next;
}lis[100005];
int main()
{
    int sta,n,k;
    cin>>sta>>n>>k;
    for(int i=0;i<n;i++)
    {
        int ad,ne,da;
        cin>>ad>>da>>ne;
        lis[ad].address=ad;
        lis[ad].data=da;
        lis[ad].next=ne;
    }
   vector<int>v[3];
    for(int i=0;i<n;i++)
    {
        if(lis[sta].data<0)
        {
            v[0].push_back(lis[sta].address);
        }
        else if(lis[sta].data<=k)
        {
            v[1].push_back(lis[sta].address);
        }
        else
        {
            v[2].push_back(lis[sta].address);
        }
        sta=lis[sta].next;
    }
    int flag=0;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<v[i].size();j++)
        {
            if(flag==0)
            {
                printf("%05d %d ",v[i][j],lis[v[i][j]].data);
                flag=1;
            }
            else
            {
                printf("%05d\n%05d %d ",v[i][j],v[i][j],lis[v[i][j]].data);
            }
        }
    }
    cout<<"-1";
    return 0;
}