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
  
 1 #include <stdio.h>
 2 #include <string>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 using namespace std;
 7 const int maxn=100010;
 8 struct node{
 9     int data;
10     int pos=0;
11     int address;
12     int next;
13     int rank;
14 }nodes[maxn];
15 vector<node> v;
16 int n,k,root;
17 bool cmp(node n1,node n2){
18     if(n1.rank<n2.rank) return true;
19     else if(n1.rank>n2.rank) return false;
20     else{
21         return n1.pos<n2.pos;
22     }
23 }
24 int main(){
25   scanf("%d %d %d",&root,&n,&k);
26   for(int i=0;i<n;i++){
27     int first,data,next;
28     scanf("%d %d %d",&first,&data,&next);
29     node tmp;
30     tmp.address = first;
31     tmp.data = data;
32     tmp.next = next;
33     if(data<0) tmp.rank=1;
34     else if(data<=k) tmp.rank=2;
35     else tmp.rank=3;
36     nodes[first]=tmp;
37     //v.push_back(tmp);
38   }
39   int j=1;
40   while(root!=-1){
41       nodes[root].pos=j;
42       v.push_back(nodes[root]);
43     root=nodes[root].next;
44     j++;
45   }
46   sort(v.begin(),v.end(),cmp);
47   for(int i=0;i<v.size();i++){
48       if(i!=v.size()-1){
49           printf("%05d %d %05d\n",v[i].address,v[i].data,v[i+1].address);
50       }
51       else {
52           printf("%05d %d -1\n",v[i].address,v[i].data);
53       }
54   }
55 }
View Code

注意点：看到只想着排序了，看大佬的方法真简单，直接三个vector，再合起来打印输出。

一开始最后第二个测试点，最后一个超时，超时是因为一开始存数据直接存在vector里，然后要得到正确的顺序每次都要遍历一遍整个vector，时间复杂度很高。而且最开始的pos其实设置的也是有问题的，直接根据输入顺序得到了，神奇的是结果居然前几个都是对的。

最后存储链表还是要用静态数组，不能用vector，找next太耗时