2017-9-17pat甲级 B-优快云博客

本文链接：https://blog.youkuaiyun.com/yu121380/article/details/78051523

本文介绍了一个链表操作问题，即按特定条件重新排列链表元素。具体而言，需要将链表中的负数节点放在前面，0到K之间的非负数节点放在中间，大于K的节点放在最后，并保持各区间内节点原有的顺序。

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B. Splitting A Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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
题目大意：给一个链表和K，遍历链表后将<0的结点先输出，再将0～k区间的结点输出，最后输出>k的结点
分析：将结点用list[10000]保存，list为node类型，node中保存结点的值value和它的next地址。list的下标就是结点的地址。将<0、0～k、>k三部分的结点地址分别保存在v[0]、v[1]、v[2]中，最后将vector中的值依次输出即可～
代码：
#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;

vector<int> v[3];

struct node
{
    int data;
    int next;
}list[100000];

int main()
{
    int start,n,k,a;
    scanf("%d%d%d",&start,&n,&k);

    for(int i=0; i<n; i++)
    {
        scanf("%d",&a);
        scanf("%d%d",&list[a].data,&list[a].next);
    }

    int p=start;
    while(p!=-1)
    {
        if(list[p].data<0)
            v[0].push_back(p);
        else if(list[p].data>=0&&list[p].data<=k)
            v[1].push_back(p);
        else
            v[2].push_back(p);
        p=list[p].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],list[v[i][j]].data);
                flag=1;
            }
            else
            {
                printf("%05d\n%05d %d ",v[i][j],v[i][j],list[v[i][j]].data);
            }
        }
        printf("-1\n");

    return 0;
}