题目：
请修改append函数，利用这个函数实现：
两个非降序链表的并集，1->2->3 和 2->3->5 并为 1->2->3->5
另外只能输出结果，不能修改两个链表的数据。
分析：
这题很简单，两个指向链表的指针，比较对应的值，并遍历
实现如下：
#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 << 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->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; }
输出如下：list1: 1->2->3
list2: 2->3->5
list1 and list2: 1->2->3->5