合并两个单链表出问题了
救命！{:10_250:}
在用来合并两个有序序列的sort()函数那里，我将两个单链表合并了（将b的结点插到a链表里），然后在这个函数里输出合并后的单链表很正常。但是我再到main函数里面输出那个单链表的时候就会输出乱七八糟的东西。
//单链表
#include<iostream>
#include<cstdlib>
#include<ctime>
using namespace std;
struct node {
int data;
node * next;
node(){
next = NULL;
}
};
//--------------------------------------------------
class list {
private:
node * head, * rear;
public:
list(){head = rear = NULL;}
void create(); //构建链表，从键盘输入数据
void create_2(); //构建链表，自动生成数据，有序
void print(); //输出链表到屏幕
~list(); //释放内存
void sort(list b); //排序
void merge(node * f1, node * e1, node * f2, node * e2);
node * get_head();
};
//--------------------------------------------------
node * list::get_head() {
return head;
}
//--------------------------------------------------
list::~list() {
node * t;
while(head) {
t = head;
head = head->next;
delete t;
}
}
//--------------------------------------------------
void list::print() {
node * t = head->next;
while(t != NULL) {
cout << t->data << " ";
t = t->next;
}
cout << endl;
}
//--------------------------------------------------
void list::create() {
head = new node;
rear = head;
node * t;
int n(0), t_data(0);
cin >> n; //数据的数量
for(int i(0); i < n; ++i) {
t = new node; //建立新结点
cin >> t_data;
t->data = t_data;
rear->next = t;
rear = rear->next;
}
}
//--------------------------------------------------
void list::create_2() {
srand(time(0));
int data(0);
head = new node;
rear = head;
node * t;
int n(0), t_data(0);
cin >> n; //数据的数量
for(int i(0); i < n; ++i) {
t = new node; //建立新结点
data += rand() % 6;
t_data = data;
t->data = t_data;
rear->next = t;
rear = rear->next;
}
}
void list::sort(list b) {
node * a1(head), * b1(b.get_head());
node * t(NULL);
b1 = b1->next;
while(b1 != NULL && a1->next != NULL) {
if(b1->data <= a1->next->data) {
t = b1->next;
b1->next = a1->next;
a1->next = b1;
b1 = t;
}
else {
a1 = a1->next;
}
print();
}
if(b1 != NULL) {
a1->next = b1;
}
cout << "last:" << endl;
print();
}
int main() {
list a, b;
a.create_2();
b.create_2();
a.print();
b.print();
a.sort(b);
cout << "b:" << endl;
cout << "a:" << endl;
a.print();
return 0;
}
本帖最后由 jhq999 于 2022-6-13 07:02 编辑
看看
运行一下没错啊？
3 3
3 4 5
4 7 11
3 4 5
3 4 4 5
3 4 4 5
3 4 4 5
3 4 4 5
last:
3 4 4 5 7 11
b:
a:
3 4 4 5 7 11 本帖最后由 jhq999 于 2022-6-13 09:41 编辑
void list::sort(list *b) {///////////////注意类实例的析构
node * a1(head), * b1(b->get_head());
node * t(NULL);
b1 = b1->next;
while(b1 != NULL && a1->next != NULL) {
if(b1->data <= a1->next->data) {
t = b1->next;
b1->next = a1->next;
a1->next = b1;
b1 = t;
}
else {
a1 = a1->next;
}
print();
}
if(b1 != NULL) {
a1->next = b1;
}
b->get_head()->next=NULL;//////////////////////如果不设为null,主程序结束时实例b会析构，会释放已经在a析构时已经释放的数据
cout << "last:" << endl;
print();
} jhq999 发表于 2022-6-13 08:49
得知答案的我心情是复杂的，但我忍不住想要高呼“jhq999 nb！”{:9_232:}
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