1133 Splitting A Linked List (25 point(s))

1133 Splitting A Linked List (25 point(s))

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

链表的操作。

1. 按照头节点指针确定在链表中的结点；

2. 按照结点元素大小3次顺序遍历； 

3. 输出（注意最后一个节点的next指针是-1）。

#include
#include
#include
using namespace std;
const int MAX = 1e5+7;
struct Node{
	int address,key,nextAdd;
};
Node node[MAX];
int start,N,K;
int main(void){
	cin>>start>>N>>K;
	int a,k,n;
	while(N--){
		cin>>a>>k>>n;
		node[a].address=a;
		node[a].key=k;
		node[a].nextAdd=n;
	}
	vector v;
	while(start!=-1){
		v.push_back(node[start]);
		start = node[start].nextAdd;
	}
	vector ans;
	for(int i=0;i=0&&v[i].key<=K) ans.push_back(v[i]);
	}
	for(int i=0;iK) ans.push_back(v[i]);
	}
	for(int i=0;i