~~~笔锋至此又怎能平淡而终,故事开始便不承认普通✌✌✌
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题目:
已知两个链表A和B分别表示两个集合,其元素递增排列,编制函数,求A与B的交集,并存放于A链表中。
解题思路:
>类似归并排序
>将不等结点释放,相等结点进行尾插
代码实现:
#include <iostream>
using namespace std;
typedef struct LNode
{
int data;
struct LNode *next;
} LNode, *LinkList;
// 头插法
void HeadInsert(LinkList &L)
{
int val = 0;
while (cin >> val)
{
LNode *s = new LNode;
s->data = val;
s->next = L->next;
L->next = s;
if (cin.get() == '\n')
{
break;
}
}
}
// 尾插法
void TailInsert(LinkList &L)
{
int val = 0;
LNode *r = L;
while (cin >> val)
{
LNode *s = new LNode;
s->data = val;
r->next = s;
r = s;
r->next = NULL;
if (cin.get() == '\n')
{
break;
}
}
}
// 遍历输出链表元素
void Print(LinkList L)
{
LNode *p = L->next;
while (p)
{
cout << p->data << '\t';
p = p->next;
}
cout << endl;
}
void PublicNode(LinkList &LA, LinkList &LB)
{
LNode *pa, *pb, *r, *q; //定义两个链表的工作结点和尾插指针
pa = LA->next;
pb = LB->next;
r = LA;
while (pa && pb)
{
//如果pa小于pb,则将其删除,同时指针后移
if (pa->data < pb->data)
{
q = pa;
pa = pa->next;
delete q;
}
else if (pa->data > pb->data)
{
q = pb;
pb = pb->next;
delete q;
}
//如果相等将pa尾插,删除pb
else
{
r->next = pa;
r = pa;
pa = pa->next;
q = pb;
pb = pb->next;
delete q;
}
}
//将剩余所有结点全部释放
while (pa)
{
q = pa;
pa = pa->next;
delete q;
}
while (pb)
{
q = pb;
pb = pb->next;
delete q;
}
r->next = NULL;
delete LB;
}
int main()
{
LinkList LA = new LNode;
LinkList LB = new LNode;
TailInsert(LA);
TailInsert(LB);
PublicNode(LA, LB);
Print(LA);
}