PAT 1133.Splitting A Linked List（25 分

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最新推荐文章于 2024-05-08 21:39:44 发布

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本文介绍了一个算法问题，即如何对单向链表进行重组，使得所有负数节点出现在非负数节点之前，并且所有在[0,K]范围内的值出现在大于K的值之前，同时保持各分类内节点原有的顺序不变。

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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 (≤103). 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 <stdio.h>
#include <vector>
using namespace std;
struct node{
    int adress,data,next;
};
int k;
int main()
{
    int begin,n;
    scanf("%d %d %d",&begin,&n,&k);
    vector<node> receive(100001);
    vector<node> list,result;
    
    for(int i=0;i<n;i++)
    {
        int address,data,next;
        scanf("%d %d %d",&address,&data,&next);
        receive[address]=node{address,data,next};
    }
    
    for(;begin!=-1;begin=receive[begin].next)
    {
        list.push_back(receive[begin]);
    }
    for(int i=0;i<list.size();i++)
    {
        if(list[i].data<0)
            result.push_back(list[i]);
    }
    for(int i=0;i<list.size();i++)
    {
        if(list[i].data>=0 && list[i].data<=k)
        {
            result.push_back(list[i]);
        }
    }
    for(int i=0;i<list.size();i++)
    {
        if(list[i].data>k)
        {
            result.push_back(list[i]);
        }
    }
    for(int i=0;i<list.size()-1;i++)
    {
        printf("%05d %d %05d\n",result[i].adress,result[i].data,result[i+1].adress);
    }
    int index = (int)list.size()-1;
    printf("%05d %d -1\n",result[index].adress,result[index].data);
    return 0;
}