1133 Splitting A Linked List（25 分）

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

思路：

给一个链表和K，遍历链表后将<0的结点先输出，再将0～k区间的结点输出，最后输出>k的结点

C++：

#include "iostream"
#include "vector"
using namespace std;
struct node
{int add,data,next;};
const int maxn=100010;
vector list,neg,sma,ori;
node accord[maxn];
void linklist(int s){
	while (s!=-1)
	{
		list.push_back(accord[s]);
		s=accord[s].next;
	}
}
int main(){
	int n,k,start;
	scanf("%d %d %d",&start,&n,&k);
	for (int i=0;i=0&&list[i].data<=k)
			sma.push_back(list[i]);
		if (list[i].data>0&&list[i].data>k)
			ori.push_back(list[i]);
	}
	for(int j=0;j