1133 Splitting A Linked List （25 分）

1133 Splitting A Linked List （25 分）

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 (≤10​5​​) which is the total number of nodes, and a positive K (≤10​3​​). 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 [−10​5​​,10​5​​], 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
思路
　　中间有一个测试点输出结果只有一个结点，导致我调了很久。

#include<iostream>
#include<string>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<string.h>
#include<algorithm>
using namespace std;

struct Node
{
    int address;
    int data;
    int next;
};

Node node[10000010];
int main()
{
    int address,n,k;
    scanf("%d%d%d",&address,&n,&k);
    for(int i=0; i<n; i++)
    {
        Node temp;
        scanf("%d%d%d",&temp.address,&temp.data,&temp.next);
        node[temp.address]=temp;
    }
    vector<Node> v1,v2,v3;
    for(int i=address; i!=-1; i=node[i].next)
    {
        if(node[i].data<0)
            v1.push_back(node[i]);
        else if(node[i].data<=k)
            v2.push_back(node[i]);
        else
            v3.push_back(node[i]);
    }
    v1.insert(v1.end(),v2.begin(),v2.end());
    v1.insert(v1.end(),v3.begin(),v3.end());
    for(int i=1;i<v1.size();i++)
    {
        printf("%05d %d %05d\n",v1[i-1].address,v1[i-1].data,v1[i].address);
        if(i==v1.size()-1)
            printf("%05d %d -1\n",v1[i].address,v1[i].data);
    }
    if(v1.size()==1)
        printf("%05d %d -1\n",v1[0].address,v1[0].data);
    return 0;
}

 

标签：node,25,10,1133,list,List,int,address,include
来源： https://www.cnblogs.com/zhanghaijie/p/10371349.html