所以我正在编写一个程序，其中用户用整数填充两个链表，现在我需要创建一个函数，创建第三个链表，其中包含来自第一个和第二个列表的值，没有重复。 这是我现在的代码：
#include <iostream>
#include <stdlib.h>
using namespace std;
struct node
{
int info;
node *next;
node (int i, node *n=NULL)
{
info = i;
next = n;
}
~node()
{
cout <<"NODE CONTAINING \"" << info << "\" WAS DELETED!" << endl;
}
};
struct list
{
node* startList1, *lastList1, *startList2, *lastList2;
int menuOption;
int nodeCount1=0, nodeCount2=0;
list() {
startList1 = NULL;
startList2 = NULL;
}
void addList1(node *p)
{
int n;
cout << "PLEASE INPUT VALUE WHICH YOU WANT IN THE NODE:";
cin >> n;
p = new node(n);
nodeCount1++;
if(startList1==NULL)
{
startList1 = lastList1 = p;
}
else
{
lastList1->next = p;
lastList1 = p;
}
}
void printList1(node *pr)
{
node *pr;
for (pr=startList1; pr!=NULL; pr=pr->next)
{
cout << pr->info << endl;
}
}
void addList2(node *q)
{
int n;
cout << "PLEASE INPUT VALUE WHICH YOU WANT IN THE NODE:";
cin >> n;
q = new node(n);
nodeCount2++;
if(startList2==NULL)
{
startList2 = lastList2 = q;
}
else
{
lastList2->next = q;
lastList2 = q;
}
}
void printList2(node *pr)
{
for (pr=startList2; pr!=NULL; pr=pr->next)
{
cout << pr->info << endl;
}
}
// this just prints first and second lists to show what is inside..
void printBoth(node *pr, node *qr)
{
cout << "Elements of the first list:" << endl;
for (pr=startList1; pr!=NULL; pr=pr->next)
{
cout << pr->info << endl;
}
cout << "Elements of the second list:" << endl;
for (pr=startList2; pr!=NULL; pr=pr->next)
{
cout << pr->info << endl;
}
}
void printMenu()
{
cout << "MENU" << endl;
cout << "(1) ADD ELEMENT LIST1." << endl;
cout << "(2) PRINT LIST1" << endl;
cout << "(3) ADD ELEMENT LIST2" << endl;
cout << "(4) PRINT LIST2" << endl;
cout << "(5) PRINT BOTH LISTS" << endl;
cout << "(6) USE MERGE FUNCTION" << endl;
cout << "(7) TO EXIT" << endl;
cin >> menuOption;
system ("cls");
};
void dragons()
{
node *temp1 = startList1;
node *temp2 = startList2;
while(temp1)
{
temp1 = startList1->next;
delete startList1;
startList1=temp1;
}
while(temp2)
{
temp2 = startList2->next;
delete startList2;
startList2=temp2;
}
};
};
int main()
{
struct node *p = NULL, *q = NULL;
list s;
s.printMenu();
node* list1;
node* list2;
node* sorting;
while(s.menuOption!=7)
{
switch (s.menuOption)
{
case 1: s.addList1(list1);
break;
case 2: s.printList1(list1);
break;
case 3: s.addList2(list2);
break;
case 4: s.printList2(list2);
break;
case 5:s.printBoth(list1, list2);
break;
case 6:s.merge();
break;
default: cout << "SOMETHING WENT WRONG!!!!" << endl;
break;
}
system ("pause");
system ("cls");
s.printMenu();
}
s.dragons();
return 0;
}
所以现在用户可以将元素输入到两个列表并查看它们，但是如何创建一个函数，它将合并那些没有重复值的列表？例如： List1 = 1,2,3,6,8; List2 = 2,4,5,7,8;
合并列表= 1,2,3,4,5,7,8; （实际上可以是任何顺序，也无关紧要）
任何建议将不胜感激！谢谢！
答案 0 :(得分：0)
这是您的代码的一个版本，但使用STL列表容器来提供链接列表，这提供了自己的功能，可以实现您想要的功能。此代码用于演示，并未针对效率进行优化，它仅显示了合并两个列表中的唯一元素的一种可能方法。 （为此，列表需要对其元素进行排序，并且需要为列表中的节点提供＆＃34;小于＆＃34;和＃34;等于＆＃34;的谓词。） （确实需要时间来学习一些STL容器和函数，但很多人会建议这样做，而不是尝试从头开始创建自己的基于原始指针的链表。）
希望这会有所帮助，或者是有意义的。
#include <iostream>
#include <stdlib.h>
#include <list>
struct node
{
int info;
node (int i)
{
info = i;
}
~node()
{
//std::cout <<"NODE CONTAINING \"" << info << "\" WAS DELETED!" << std::endl;
}
};
bool nodeLess(node n1, node n2)
{
return n1.info < n2.info;
}
bool nodeEqual(node n1, node n2)
{
return n1.info == n2.info;
}
struct mylist
{
int menuOption;
std::list<node> list1;
std::list<node> list2;
void addList(std::list<node>& l)
{
int x;
std::cout << "PLEASE INPUT VALUE WHICH YOU WANT IN THE NODE: ";
std::cin >> x;
node n(x);
l.push_back(n);
}
void printList(const std::list<node>& l)
{
for (std::list<node>::const_iterator it = l.cbegin(); it != l.cend(); ++it)
{
std::cout << it->info << std::endl;
}
}
void addList1() { addList(list1); }
void addList2() { addList(list2); }
void printList1() { printList(list1); }
void printList2() { printList(list2); }
// this just prints first and second lists to show what is inside..
void printBoth()
{
std::cout << "Elements of the first list:" << std::endl;
printList1();
std::cout << "Elements of the second list:" << std::endl;
printList2();
}
void simpleMerge()
{
std::list<node> merged;
merged.insert(merged.end(), list1.begin(), list1.end());
merged.insert(merged.end(), list2.begin(), list2.end());
std::cout << "CONTENTS OF LIST1 THEN LIST2: " << std::endl;
printList(merged);
}
void uniqueSortMerge()
{
std::list<node> sorted1(list1.begin(), list1.end());
std::list<node> sorted2(list2.begin(), list2.end());
sorted1.sort(nodeLess);
sorted2.sort(nodeLess);
sorted1.unique(nodeEqual);
sorted2.unique(nodeEqual);
std::list<node> merged;
std::merge(sorted1.begin(), sorted1.end(),
sorted2.begin(), sorted2.end(),
std::back_inserter(merged),
nodeLess);
std::cout << "UNIQUE CONTENTS OF LIST1 AND LIST2 SORTED AND MERGED: " << std::endl;
printList(merged);
}
void printMenu()
{
std::cout << "MENU" << std::endl;
std::cout << "(1) ADD ELEMENT LIST1." << std::endl;
std::cout << "(2) PRINT LIST1" << std::endl;
std::cout << "(3) ADD ELEMENT LIST2" << std::endl;
std::cout << "(4) PRINT LIST2" << std::endl;
std::cout << "(5) PRINT BOTH LISTS" << std::endl;
std::cout << "(6) USE SIMPLE MERGE FUNCTION" << std::endl;
std::cout << "(7) USE UNIQUE, SORT AND MERGE FUNCTION" << std::endl;
std::cout << "(8) TO EXIT" << std::endl;
std::cin >> menuOption;
system ("cls");
};
void dragons()
{
list1.clear();
list2.clear();
};
};
int main()
{
mylist s;
do
{
s.printMenu();
switch (s.menuOption)
{
case 1:
s.addList1();
break;
case 2:
s.printList1();
break;
case 3:
s.addList2();
break;
case 4:
s.printList2();
break;
case 5:
s.printBoth();
break;
case 6:
s.simpleMerge();
break;
case 7:
s.uniqueSortMerge();
break;
case 8:
break;
default:
std::cout << "SOMETHING WENT WRONG!!!!" << std::endl;
break;
}
system ("pause");
system ("cls");
} while(s.menuOption != 8);
s.dragons();
return 0;
}