1097. Deduplication on a Linked List (25)【链表】——PAT (Advanced Level) Practise

题目信息

1097. Deduplication on a Linked List (25)

时间限制300 ms

内存限制65536 kB

代码长度限制16000 B

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (<= 105) which is the total number of nodes. 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 Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 10^4, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 5

99999 -7 87654

23854 -15 00000

87654 15 -1

00000 -15 99999

00100 21 23854

Sample Output:

00100 21 23854

23854 -15 99999

99999 -7 -1

00000 -15 87654

87654 15 -1

解题思路

链表标记连接节点

AC代码

#include <cstdio>
#include <set>
#include <map>
#include <vector>
using namespace std;
set<int> st;
map<int, pair<int, int> > mp;
vector<int> v;
int main()
{
int first, n, a, b, k;
scanf("%d%d", &first, &n);
for (int i = 0; i < n; ++i){
scanf("%d%d%d", &a, &k, &b);
mp[a] = make_pair(k, b);
}
int p = first;
while (p != -1){
printf("%05d %d ", p, mp[p].first);
st.insert(abs(mp[p].first));
p = mp[p].second;
while (p != -1 && st.find(abs(mp[p].first)) != st.end()){
v.push_back(p);
p = mp[p].second;
}
if (~p) printf("%05d\n", p);
}
printf("-1\n");
if (!v.empty()){
printf("%05d %d ", v[0], mp[v[0]].first);
for (int i = 1; i < v.size(); ++i){
printf("%05d\n%05d %d ", v[i], v[i], mp[v[i]].first);
}
printf("-1\n");
}
return 0;
}