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 (≤103). 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 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 ",v[i].address,v[i].data,v[i+1].address); 50 } 51 else { 52 printf("%05d %d -1 ",v[i].address,v[i].data); 53 } 54 } 55 }
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