PAT甲级 1133. Splitting A Linked List (25)

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 (<=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 [-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

三次遍历所有输入即可保证相对顺序与原先一样，然后依次把符合条件的存入结果链表中。

#include
#include
using namespace std;

struct node {
int data, next;
node() { data = 1000000; next = -1; }
};

string printzero(int num)
{
if (num == 0)
return "0000";
if (num == -1)
return "";
string s = "";
while (num<10000)
{
s.append("0");
num *= 10;
}
return s;
}

int main()
{
int start, N, K, resultnum = 0;
node * list = new node[100002];
node * result = new node[100002];
cin >> start >> N >> K;
for (int i = 0; i> adress;
cin >> list[adress].data >> list[adress].next;
}
int lastadress = -1, adress = start, resultstart = -1;
while (adress != -1)
{
if (list[adress].data<0)
{
result[adress] = list[adress];
resultnum++;
if (resultnum == 1)
resultstart = adress;
else
result[lastadress].next = adress;
lastadress = adress;
}
adress = list[adress].next;
}
adress = start;
while (adress != -1)
{
if (list[adress].data >= 0 && list[adress].data <= K)
{
result[adress] = list[adress];
resultnum++;
if (resultnum == 1)
resultstart = adress;
else
result[lastadress].next = adress;
lastadress = adress;
}
adress = list[adress].next;
}
adress = start;
while (adress != -1)
{
if (list[adress].data>K)
{
result[adress] = list[adress];
resultnum++;
if (resultnum == 1)
resultstart = adress;
else
result[lastadress].next = adress;
lastadress = adress;
}
adress = list[adress].next;
}
result[lastadress].next = -1;
adress = resultstart;

while (adress != -1)
{
cout << printzero(adress) << adress << " " << result[adress].data << " " << printzero(result[adress].next) << result[adress].next << endl;
adress = result[adress].next;
}
return 0;
}