函数题之两个有序链表序列的合并

题目要求

本题要求实现一个函数，将两个链表表示的递增整数序列合并为一个非递减的整数序列。
函数原型为

List Merge( List L1, List L2 );

代码

List Merge( List L1, List L2 ){
    List p=L1->Next;
    List q=L2->Next;
    List L=(List)malloc(sizeof(struct Node));//创建头结点
    L->Next=NULL;
    List k=L;//尾结点
    List newnode;//新节点
    while(p&&q){
        newnode=(List)malloc(sizeof(struct Node));
        newnode->Next=NULL;
        if(p->Data<=q->Data){
            newnode->Data=p->Data;
            //q=L1->Next;
            p=p->Next;
            k->Next=newnode;
            k=k->Next;
        }
        else if(p->Data>q->Data){
            newnode->Data=q->Data;
            q=q->Next;
            k->Next=newnode;
            k=k->Next;
        }
    }
    if(p!=NULL) k->Next=p;
    if(q!=NULL)k->Next=q;
    L1->Next=NULL;
    L2->Next=NULL;
    return L;
}

复习感悟

复写的时候才注意到不能使用新节点，其实刚开始写的时候也想到了要直接rear->next=q;但是以为这样不行，后来在vs中调试跑了一下，明白了过程，对链表有了更深的了解
题目要求的合并代码：

list merge(list l1, list l2) {
	list l3=new Node;
	l3->next = NULL;
	list rear = l3;
	list q = l1->next;
	list p = l2->next;
	while (q && p) {
		if (q->data < p->data) {
			rear->next = q;
			q = q->next;
			rear = rear->next;
		}
		else {
			rear->next = p;
			p = p->next;
			rear = rear->next;
		}
	}
	if (q) rear->next = q;
	if (p) rear->next = p;
	l1->next = NULL;
	l2->next = NULL;
	return l3;
}

总体代码：

#include<iostream>
using namespace std;
struct Node {
	int data;
	Node* next;
};
typedef Node* list;
list read() {
	int n;
	cin >> n;
	list head=new Node;
	head->next = NULL;
	list rear = head;
	list newnode;
	while (n--) {
		newnode = new Node;
		cin >> newnode->data;
		newnode->next = NULL;
		rear->next = newnode;
		rear = rear->next;
	}
	return head;
}
void printList(list l) {
	list q = l->next;
	if (!q) cout << "NULL";
	while (q) {
		cout << q->data<<" ";
		q = q->next;
	}
	cout << endl;
}
list merge(list l1, list l2) {
	list l3=new Node;
	l3->next = NULL;
	list rear = l3;
	list q = l1->next;
	list p = l2->next;
	while (q && p) {
		if (q->data < p->data) {
			rear->next = q;
			q = q->next;
			rear = rear->next;
		}
		else {
			rear->next = p;
			p = p->next;
			rear = rear->next;
		}
	}
	if (q) rear->next = q;
	if (p) rear->next = p;
	l1->next = NULL;
	l2->next = NULL;
	return l3;
}
int main() {
	list l1 = read();
	list l2 = read();
	list l3 = merge(l1, l2);
	printList(l3);
	printList(l1);
	printList(l2);
	return 0;
}

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