PAT_1133

1133 Splitting A Linked List

题目描述

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:

1
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:

1
2
3
4
5
6
7
8
9
10
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:

1
2
3
4
5
6
7
8
9
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

可以发现,将所有节点分为三部分,首先是小于0,第二部分为大于等于0且小于等于k,第三部分为大于k的,我们需要将相同部分交换到相邻位置,可以直接使用三个queue来进行模拟即可

程序

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 1e5+5;
struct Node
{
int add,date,next;
}node[maxn];
int add,N,K;
int mapp[maxn];
queue<int> negative,small,last;

int main()
{
scanf("%d%d%d",&add,&N,&K);
for(int i =0 ;i < N;i ++)
{
scanf("%d%d%d",&node[i].add,&node[i].date,&node[i].next);
mapp[node[i].add] = i;
}
int k = add;
while(k != -1)
{
if(node[mapp[k]].date < 0)
negative.push(node[mapp[k]].add);
else
{
if(node[mapp[k]].date <= K)
small.push(node[mapp[k]].add);
else
last.push(node[mapp[k]].add);
}
k = node[mapp[k]].next;
}
Node tmp;
bool flag = false;;
while(!negative.empty())
{
tmp = node[mapp[negative.front()]];negative.pop();
if(!flag)
printf("%05d %d",tmp.add,tmp.date),flag = true;
else
printf(" %05d\n",tmp.add),printf("%05d %d",tmp.add,tmp.date);
}
while(!small.empty())
{
tmp = node[mapp[small.front()]];small.pop();
if(!flag)
printf("%05d %d",tmp.add,tmp.date),flag = true;
else
printf(" %05d\n",tmp.add),printf("%05d %d",tmp.add,tmp.date);
}
while(!last.empty())
{
tmp = node[mapp[last.front()]];last.pop();
if(!flag)
printf("%05d %d",tmp.add,tmp.date),flag = true;
else
printf(" %05d\n",tmp.add),printf("%05d %d",tmp.add,tmp.date);
}
printf(" -1\n");
return 0;
}
-------------本文结束感谢您的阅读-------------
显示 Gitment 评论