PAT 1133 Splitting A Linked List

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

#include<iostream> //偏简单， 看清题目
#include<vector>
using namespace std;
struct node{
    int val;
    int next;
};
int main(){
    int First, n, k, t=-1, cnt=0;
    cin>>First>>n>>k;
    vector<node> data(100000);
    vector<int> negative, posl, posr, ans;
    for(int i=0; i<n; i++){
        int d, v, next;
        cin>>d;
        cin>>data[d].val>>data[d].next;
    }
    while(First!=-1){
        if(data[First].val<0)
            negative.push_back(First);
        else if(data[First].val<=k)
            posl.push_back(First);
        else if(data[First].val>k)
            posr.push_back(First);
        First=data[First].next;
    }
    for(int i=0; i<negative.size(); i++)
        ans.push_back(negative[i]);
    for(int i=0; i<posl.size(); i++)
        ans.push_back(posl[i]);
    for(int i=0; i<posr.size(); i++)
        ans.push_back(posr[i]);
    for(int i=0; i<ans.size(); i++)
        if(i!=ans.size()-1)
            printf("%05d %d %05d\n", ans[i], data[ans[i]].val, ans[i+1]);
        else
            printf("%05d %d -1", ans[i], data[ans[i]].val);
    return 0;
} 

原文地址：https://www.cnblogs.com/A-Little-Nut/p/9506883.html