标签：nes sample fir link represent clu node 难度 val
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.
Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.
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 1, and Next is the position of the next node.
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.
00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854
00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1
题意：
本质上是一道链表的题目，但是改变了链表的表示方法，增加了难度。
思路：
每一个结点的头指针是不会改变的，只会改变其尾指针。
Code：
#include<iostream>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<string>
using namespace std;
struct Node {
string addr;
int key;
string next;
};
map<string, Node> m;
set<int> s;
queue<Node> banch;
string findNext(string start) {
auto it = m.find(start);
int k = abs(it->second.key);
if (it->second.next == "-1") {
if (s.find(k) != s.end()) {
banch.push(it->second);
return "-1";
} else return start;
}
if (s.find(k) != s.end()) {
banch.push(it->second);
} else {
return start;
}
return findNext(it->second.next);
}
int main() {
string start;
int n;
cin >> start >> n;
vector<Node> v1, v2;
Node node;
for (int i = 0; i < n; ++i) {
int key;
string addr, next;
cin >> addr >> key >> next;
node = {addr, key, next};
m.insert({addr, node});
}
queue<Node> master;
for (int i = 0; i < n; ++i) {
auto it = m.find(start);
node = it->second;
s.insert(abs(node.key));
node.next = findNext(node.next);
master.push(node);
start = node.next;
if (start == "-1") break;
}
while(!master.empty()) {
node = master.front();
if (master.size() == 1) cout << node.addr << " " << node.key << " " << "-1" << endl;
else cout << node.addr << " " << node.key << " " << node.next << endl;
master.pop();
}
Node next_node;
while(!banch.empty()) {
node = banch.front();
if (banch.size() == 1) {
banch.pop();
cout << node.addr << " " << node.key << " " << "-1" << endl;
} else {
banch.pop();
cout << node.addr << " " << node.key << " " << banch.front().addr << endl;
}
}
return 0;
}
可能是因为随着数据量的增大，会使得递归调用的层数很深，提交的时候有两组数据显示“Segmentation Fault”。
经过改进的代码消除一个“Segmentation Fault”错误，主要是做这道题的思路变了很多。
#include<iostream>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<cstdio>
using namespace std;
typedef struct Node *node;
struct Node {
int addr;
int key;
int next;
node nextnode;
Node(int s1, int k, int s2) : addr(s1), key(k), next(s2), nextnode(NULL) {}
};
int main() {
int n;
int start;
scanf("%d%d", &start, &n);
node Head1 = (node)malloc(sizeof(struct Node));
node Head2 = (node)malloc(sizeof(struct Node));
node prt1, prt2, pre;
prt1 = Head1;
prt2 = Head2;
map<int, node> m;
set<int> s;
node num, temp;
for (int i = 0; i < n; ++i) {
int key;
int addr, next;
scanf("%d%d%d", &addr, &key, &next);
num = new Node(addr, key, next);
m.insert({addr, num});
}
for (int i = 0; i < n; ++i) {
temp = m.find(start)->second;
prt1->nextnode = temp;
prt1 = prt1->nextnode;
start = prt1->next;
}
prt1 = Head1->nextnode;
pre = Head1;
while (prt1) {
int k = abs(prt1->key);
if (s.find(k) == s.end()) {
prt1 = prt1->nextnode;
pre = pre->nextnode;
s.insert(k);
} else {
prt2->nextnode = prt1;
prt2 = prt2->nextnode;
pre->nextnode = prt1->nextnode;
prt1 = prt1->nextnode;
}
}
prt1 = Head1->nextnode;
prt2 = Head2->nextnode;
while (prt1) {
if (prt1->nextnode == NULL)
printf("%05d %d -1\n", prt1->addr, prt1->key);
else
printf("%05d %d %05d\n", prt1->addr, prt1->key, prt1->nextnode->addr);
prt1 = prt1->nextnode;
}
while (prt2) {
if (prt2->nextnode == NULL)
printf("%05d %d -1\n", prt2->addr, prt2->key);
else
printf("%05d %d %05d\n", prt2->addr, prt2->key, prt2->nextnode->addr);
prt2 = prt2->nextnode;
}
return 0;
}
1097 Deduplication on a Linked List
标签：nes sample fir link represent clu node 难度 val
原文地址：https://www.cnblogs.com/ruruozhenhao/p/12609814.html
