1133. Splitting A Linked List (25)

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], andNext 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 1023333 10 2777700000 0 9999900100 18 1230968237 -6 2333333218 -4 0000048652 -2 -199999 5 6823727777 11 4865212309 7 33218

Sample Output:

33218 -4 6823768237 -6 4865248652 -2 1230912309 7 0000000000 0 9999999999 5 2333323333 10 0010000100 18 2777727777 11 -1

#include <iostream>#include <vector>#include <cstdio>#include <algorithm>using namespace std;struct xnode{    int add,val,next;};xnode node[100000];vector<xnode> p,q,r,ans;int head,n,k;int main(){    cin>>head>>n>>k;    for(int i=0;i<n;++i)    {        int add,val,next;        cin>>add>>val>>next;        node[add].add=add;        node[add].val=val;        node[add].next=next;    }    int x=head;    while(x!=-1)    {        if(node[x].val<0) p.push_back(node[x]);        else if(node[x].val>=0&&node[x].val<=k) q.push_back(node[x]);        else r.push_back(node[x]);        x=node[x].next;    }    ans=p;    for(auto ln:q) ans.push_back(ln);    for(auto ln:r) ans.push_back(ln);    int al=ans.size();    for(int i=0;i<al-1;++i)        printf("%05d %d %05d\n",ans[i].add,ans[i].val,ans[i+1].add);    printf("%05d %d -1\n",ans[al-1].add,ans[al-1].val);    return 0;}