两个单链表合并问题

最新推荐文章于 2025-06-06 01:13:22 发布

雄鹰展翅2017

最新推荐文章于 2025-06-06 01:13:22 发布

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分类专栏： c-c++

本文链接：https://blog.youkuaiyun.com/qq_28959531/article/details/78155215

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本文介绍了一种合并两个有序单链表的方法，提供了非递归和递归两种实现方式，并通过测试代码验证了其正确性。

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链表的结点如下所示，已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序。

struct Node
{
    int data ;
    Node *next ;
};
typedef struct Node Node ;

单链表合并代码，非递归方式

Node * Merge(Node *head1 , Node *head2)
{
    if ( head1 == NULL )   //head1为空 
        return head2 ;
    if ( head2 == NULL)    //head2为空 
        return head1 ;

    Node *head = NULL ;
    Node *p1 = NULL;
    Node *p2 = NULL;

    //数据链表第一个结点小的做首结点 
    if ( head1->data < head2->data )   
    {
        head = head1 ;
        p1 = head1->next;
        p2 = head2 ;
    }
    else
    {
        head = head2 ;
        p2 = head2->next ;
        p1 = head1 ;
    }

    Node *pcurrent = head ;
    //当p1移到末尾或p2移到末尾 
    while ( p1 != NULL && p2 != NULL)
    {
        if ( p1->data <= p2->data )
        {
            pcurrent->next = p1 ;
            pcurrent = p1 ;
            p1 = p1->next ;
        }
        else
        {
            pcurrent->next = p2 ;
            pcurrent = p2 ;
            p2 = p2->next ;
        }
    }
    //p1没有移到末尾，将p1添加到新链表后面 
    if ( p1 != NULL )
        pcurrent->next = p1 ;
    //p2没有移到末尾，将p2添加到新链表后面      
    if ( p2 != NULL )
        pcurrent->next = p2 ;

    return head ;
}

单链表合并代码，递归方式

Node * MergeRecursive(Node *head1 , Node *head2)
{
    if ( head1 == NULL )
        return head2 ;
    if ( head2 == NULL)
        return head1 ;
    Node *head = NULL ;
    if ( head1->data < head2->data )
    {
        head = head1 ;
        head->next = MergeRecursive(head1->next,head2);
    }
    else
    {
        head = head2 ;
        head->next = MergeRecursive(head1,head2->next);
    }
    return head ;
}

测试代码如下所示

#include <iostream>
using namespace std;

struct Node
{
    int data ;
    Node *next ;
};
typedef struct Node Node ;

//创建链表节点
Node* CreateNode()
{
    Node *head = NULL; 
    Node *pnew = NULL;
    Node *tail = NULL;
    int d; 
    while(1)
    {
        cin >> d;
        if (d==0)
            break;
        pnew = new Node;
        pnew->data = d;
        pnew->next = NULL;
        if(head == NULL)
        {
            head = pnew;
            tail = pnew;
        }
        else
        {
            tail->next = pnew;
            tail = pnew;
        }
   }
    return head;
}

void PrintList(Node *head)
{
    if (head== NULL)
        return ;
    while(head)
    {
        cout << head->data << "  ";
        head = head->next;
    }
    cout << endl;
}
//单链表合并，非递归方式 
Node * Merge(Node *head1 , Node *head2)
{
    if ( head1 == NULL )   //head1为空 
        return head2 ;
    if ( head2 == NULL)    //head2为空 
        return head1 ;

    Node *head = NULL ;
    Node *p1 = NULL;
    Node *p2 = NULL;

    //数据链表第一个结点小的做首结点 
    if ( head1->data < head2->data )   
    {
        head = head1 ;
        p1 = head1->next;
        p2 = head2 ;
    }
    else
    {
        head = head2 ;
        p2 = head2->next ;
        p1 = head1 ;
    }

    Node *pcurrent = head ;
    //当p1移到末尾或p2移到末尾 
    while ( p1 != NULL && p2 != NULL)
    {
        if ( p1->data <= p2->data )
        {
            pcurrent->next = p1 ;
            pcurrent = p1 ;
            p1 = p1->next ;
        }
        else
        {
            pcurrent->next = p2 ;
            pcurrent = p2 ;
            p2 = p2->next ;
        }
    }
    //p1没有移到末尾，将p1添加到新链表后面 
    if ( p1 != NULL )
        pcurrent->next = p1 ;
    //p2没有移到末尾，将p2添加到新链表后面      
    if ( p2 != NULL )
        pcurrent->next = p2 ;

    return head ;
}

//单链表合并，递归方式 
Node * MergeRecursive(Node *head1 , Node *head2)
{
    if ( head1 == NULL )
        return head2 ;
    if ( head2 == NULL)
        return head1 ;
    Node *head = NULL ;
    if ( head1->data < head2->data )
    {
        head = head1 ;
        head->next = MergeRecursive(head1->next,head2);
    }
    else
    {
        head = head2 ;
        head->next = MergeRecursive(head1,head2->next);
    }
    return head ;
}

int main(int argc, char ** argv)
{
    Node *head1 = CreateNode();
    Node *head2 = CreateNode();
    cout << "合并：" << endl; 
    PrintList(Merge(head1,head2));

    Node *head3 = CreateNode();
    Node *head4 = CreateNode();
    cout << "合并：" << endl; 
    PrintList(MergeRecursive(head3,head4));
    return 0;
}