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], 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 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

遍历三次链表，依次存储

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<set>#include<queue>#include<cstring>#include<map>using namespace std;typedef struct node{int num,next;}node;typedef struct link{int num,ad;}link;int main(){int first,n,k;cin>>first>>n>>k;node no[99999+10];for(int i=0;i<n;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);no[a].num=b;no[a].next=c;}link l[100000];int f=first;int cnt=0;while(1){if(no[f].num<0){l[cnt].num=no[f].num;l[cnt].ad=f;cnt++;}f=no[f].next;if(f==-1) break;}f=first;while(1){if(no[f].num>=0&&no[f].num<=k){l[cnt].num=no[f].num;l[cnt].ad=f;cnt++;}f=no[f].next;if(f==-1) break;}f=first;while(1){if(no[f].num>k){l[cnt].num=no[f].num;l[cnt].ad=f;cnt++;}f=no[f].next;if(f==-1) break;}for(int i=0;i<cnt-1;i++){printf("%.5d %d %.5d\n",l[i].ad,l[i].num,l[i+1].ad); }printf("%.5d %d -1\n",l[cnt-1].ad,l[cnt-1].num);return 0;}