剑指offer_16_合并两个排序的链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。#include <iostream>using namespace std;
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
void append(ListNode * &phead, int num)
{
if (phead == NULL)
{
phead = new ListNode(num);
}
else
{
ListNode* pnode = phead;
while (pnode->next != NULL)
{
pnode = pnode->next;
}
pnode->next = new ListNode(num);
}
}
class Solution {//非递归
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
ListNode* pHead = NULL, *rt = NULL;
if (pHead1 == NULL)return pHead2;
if (pHead2 == NULL)return pHead1;
if (pHead1->val<pHead2->val){
rt = pHead = pHead1;
pHead1 = pHead1->next;
}
else {
rt = pHead = pHead2;
pHead2 = pHead2->next;
}
while (1){
if (pHead1 == NULL){
pHead->next = pHead2;
break;
}
if (pHead2 == NULL){
pHead->next = pHead1;
break;
}
if (pHead1->val<pHead2->val){
pHead->next = pHead1;
pHead = pHead->next;
pHead1 = pHead1->next;
}
else{
pHead->next = pHead2;
pHead = pHead->next;
pHead2 = pHead2->next;
}
}
return rt;
}
};
class Solution1 {//递归实现
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
ListNode* pHead = NULL, *rt = NULL;
if (pHead1 == NULL)return pHead2; 
if (pHead2 == NULL)return pHead1; 
if (pHead1->val<pHead2->val){
pHead = pHead1;
pHead1 = pHead1->next;
}
else {
   pHead = pHead2;
   pHead2 = pHead2->next;
}
pHead->next = Merge(pHead1,pHead2);
return pHead;
}
};
int main()
{
ListNode *chain1=NULL,*chain2 = NULL;
for (int i = 0; i < 15; i++)
{
if (i&1)append(chain1, i);
else append(chain2, i);
}
Solution mys;
ListNode *phead = mys.Merge(chain1, chain2);
while (phead != NULL){
cout << phead->val << " ";
phead = phead->next;
}
cout << "end" << endl;
system("pause");
}