Microsoft face question resolution: Please modify the append function, using function implementation (linked list)

Topic:

Please modify the Append function to use this function to implement:

The set of two non-descending linked lists, 1->2->3 and 2->3->5, and 1->2->3->5

In addition, you can only output the results, you can not modify the data of two linked lists.

Analysis:

This question is very simple, two pointers to the linked list, compare the corresponding values, and traverse

Implemented as follows:

#include <iostream> using namespace std;  
        struct node{Node (int _v = 0): Value (_v), Next (NULL) {} int value;  
Node *next;  
};  
        void Printnodes (node* N1, node* n2) {node* P1 = n1;  
        node* P2 = n2;  
                while (P1!= null && p2!= null) {if (P1->value < P2->value)  
                        {cout << p1->value << "->";  
                P1 = p1->next; else if (P1->value > P2->value) {cout << p2  
                        ->value << "->";  
                P2 = p2->next; else {if (P1->next = null && P2->next = NULL  
                        ) cout << p2->value << Endl;
                   Else             cout << p2->value << "->";  
                        P2 = p2->next;  
                P1 = p1->next;  
                        }} while (P1!= null) {if (P1->next = null)  
                cout << p1->value << Endl;  
                else cout << p1->value << "->";  
        P1 = p1->next; } while (P2!= null) {if (P2->next = null) cout <&lt ;  
                P2->value << Endl;  
                else cout << p2->value << "->";  
        P2 = p2->next;  
        int main () {Node N1 (1), N2 (2), N3 (3), N4 (2), N5 (3), N6 (5);  
        node* Node1 = &n1;  
        Node1->next = &n2;  
        Node1->next->next = &n3;  
        node* Node2 = &n4; node2-&Gt;next = &n5;  
      
        Node2->next->next = &n6;  
        cout << "List1:";  
        Printnodes (Node1, NULL);  
        cout << "List2:";  
        Printnodes (Node2, NULL);  
        cout << "List1 and List2:";  
        Printnodes (Node1, Node2);  
return 0; }

The output is as follows:

List1:1->2->3

List2:2->3->5

List1 and List2:1->2->3->5

Author: csdn Blog hhh3h

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