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

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本文介绍了一种链表操作算法，旨在将链表中的元素按照特定条件进行分组：所有负数出现在非负数之前，所有在[0,K]范围内的数值出现在大于K的数值前，同时保持各分类内元素顺序不变。通过三次遍历链表，实现对链表的有效重组。

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 (≤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

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<iostream>
#include<vector>
#include<algorithm>
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<Node> v;
	while(start!=-1){
		v.push_back(node[start]);
		start = node[start].nextAdd;
	}
	vector<Node> ans;
	for(int i=0;i<v.size();i++){
		if(v[i].key<0) ans.push_back(v[i]);
	}
	for(int i=0;i<v.size();i++){
		if(v[i].key>=0&&v[i].key<=K) ans.push_back(v[i]);
	}
	for(int i=0;i<v.size();i++){
		if(v[i].key>K) ans.push_back(v[i]);
	}
	for(int i=0;i<ans.size()-1;i++) printf("%05d %d %05d\n",ans[i].address,ans[i].key,ans[i+1].address);
	printf("%05d %d -1\n",ans.back().address,ans.back().key);
	return 0;
}