剑指Offer--017-合并两个排序的链表

链接

题意

常规写法

#include <iostream>

using namespace std;

//  调试开关
#define __tmain main

#ifdef __tmain

#define debug cout

#else

#define debug 0 && cout

#endif // __tmain

#ifdef __tmain
struct ListNode
{
public :
int val;
struct ListNode *next;
//  ListNode(int x)
//  :val(x), next(NULL)
//    {
//  }
};

#endif // __tmain

class Solution
{
public:
ListNode* Merge(ListNode *pLeft, ListNode *pRight)
{

if(pLeft == NULL)
{
debug <<"left list is NULL" <<endl;
return pRight;
}
else if(pRight == NULL)
{
debug <<"right list is NULL" <<endl;
return pLeft;
}

ListNode *left = pLeft;
ListNode *right = pRight;

//  先生成头结点
ListNode *head = NULL;
if(left->val < right->val)
{
head = left;
left = left->next;
debug <<head->val <<"in list" <<endl;
}
else
{
head = right;
right = right->next;
debug <<head->val <<"in list" <<endl;
}

//  遍历两个链表
ListNode *curr = head;

while(left != NULL && right != NULL)
{
//  每次找到一个小的加入新的链表
debug <<"left = " <<left->val <<", right = " <<right->val <<endl;

if(left->val < right->val)
{
debug <<left->val <<"in list" <<endl;

curr->next = left;
curr = curr->next;

left = left->next;
}
else
{
debug <<right->val <<"in list" <<endl;

curr->next = right;
curr = curr->next;

right = right->next;
}
}

//  处理较长链表的剩余部分
if(left != NULL)
{
curr->next = left;
}
else
{
curr->next = right;
}

return head;

}
};

int __tmain( )
{
ListNode left, right[3];

left.val = 5;
left.next = NULL;

right[0].val = 1;
right[0].next = &right[1];
right[1].val = 2;
right[1].next = &right[2];
right[2].val = 4;
right[2].next = NULL;

Solution solu;
ListNode *head = solu.Merge(&left, right);
while(head != NULL)
{
cout <<head->val;
head = head->next;
}

return 0;
}

递归实现

class Solution
{
public:
ListNode* Merge(ListNode *pLeft, ListNode *pRight)
{

if(pLeft == NULL)
{
debug <<"left list is NULL" <<endl;
return pRight;
}
else if(pRight == NULL)
{
debug <<"right list is NULL" <<endl;
return pLeft;
}

ListNode *head = NULL;

if(pLeft->val < pRight->val)
{
head = pLeft;
head->next = Merge(pLeft->next, pRight);
}
else
{
head = pRight;
head->next = Merge(pLeft, pRight->next);
}

return head;
}
};