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 (<=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 [-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
题目大意：给出一些节点信息，按照节点在链表中原来的位置，把链表分为三段，一段的值小于零， 一段的值在[0,k]之间， 另一段的值大于k；
思路：用两个数组保存节点的值（value）和指向下一节点的指针(link); 根据节点的值把节点添加到不同的链表中； 然后在把所有的节点按照顺序添加到一个链表中；
也可以不用把所有的节点添加到一个链表中，但是输出信息的时候，要判断的情况态度了，容易出现段错误
1 #include<iostream>
2 #include<vector>
3 using namespace std;
4 struct node{
5 int addr, val, next;
6 };
7
8 int main(){
9 int root, n, k, i;
10 cin>>root>>n>>k;
11 vector<node> v1, v2, v3;
12 vector<int> link(100000, -1), value(100000);
13 for(i=0; i<n; i++){
14 int addr, val, next;
15 scanf("%d%d%d", &addr, &val, &next);
16 link[addr]=next;
17 value[addr]=val;
18 }
19 while(true){
20 if(root==-1) break;
21 node nnode; nnode.addr=root; nnode.val=value[root];
22 if(value[root]<0) v1.push_back(nnode);
23 else if(value[root]<=k) v2.push_back(nnode);
24 else v3.push_back(nnode);
25 root=link[root];
26 }
27 for(i=0; i<v2.size(); i++) v1.push_back(v2[i]);
28 for(i=0; i<v3.size(); i++) v1.push_back(v3[i]);
29 for(i=0; i<v1.size()-1; i++)printf("%05d %d %05d\n", v1[i].addr, v1[i].val, v1[i+1].addr);
30 printf("%05d %d %d\n", v1[v1.size()-1].addr, v1[v1.size()-1].val, -1);
31 return 0;
32 }
原文地址：https://www.cnblogs.com/mr-stn/p/9179407.html