合并单链表，输出单链表中间元素，判断是否有环等

1. 合并两个有序的单链表成一个有序的单链表
方法分为递归实现与非递归实现，两种方法都不额外开辟内存空间

链表的数据结构在本博客的单链表逆转，约瑟夫环等

递归实现：

//递归实现合并两个有序单链表  
LinkNode* merge_list(LinkNode *pHead1, LinkNode *pHead2) 
{ 
    if(pHead1==NULL) 
        return pHead2; 
    if(pHead2==NULL) 
        return pHead1; 
    if(pHead1==NULL && pHead2==NULL) 
        return NULL; 
    LinkNode *pMergedHead=NULL; 
    if(pHead1->value<pHead2->value) 
    { 
        pMergedHead=pHead1; 
        pMergedHead->next = merge_list(pHead1->next, pHead2); 
    } 
    else  
    { 
        pMergedHead=pHead2; 
        pMergedHead->next=merge_list(pHead1, pHead2->next); 
    } 
    return pMergedHead; 
} 

//递归实现合并两个有序单链表
LinkNode* merge_list(LinkNode *pHead1, LinkNode *pHead2)
{
 if(pHead1==NULL)
  return pHead2;
 if(pHead2==NULL)
  return pHead1;
 if(pHead1==NULL && pHead2==NULL)
  return NULL;
 LinkNode *pMergedHead=NULL;
 if(pHead1->value<pHead2->value)
 {
  pMergedHead=pHead1;
  pMergedHead->next = merge_list(pHead1->next, pHead2);
 }
 else
 {
  pMergedHead=pHead2;
  pMergedHead->next=merge_list(pHead1, pHead2->next);
 }
 return pMergedHead;
}

非递归实现：

//非递归实现合并两个有序单链表(不额外开辟空间)  
LinkNode* non_merge_list(LinkNode *pHead1, LinkNode *pHead2) 
{ 
    if(pHead1==NULL) 
        return pHead2; 
    if(pHead2==NULL) 
        return pHead1; 
    if(pHead1==NULL && pHead2==NULL) 
        return NULL; 
    LinkNode *pMergedHead = NULL; 
    LinkNode *q=NULL; 
    if(pHead1->value<pHead2->value) 
    { 
        pMergedHead=pHead1; 
        pHead1=pHead1->next; 
    } 
    else 
    { 
        pMergedHead=pHead2; 
        pHead2=pHead2->next; 
    }    
    q=pMergedHead; 
    while(pHead1 && pHead2) 
    { 
        if(pHead1->value<pHead2->value) 
        { 
            q->next=pHead1; 
            pHead1=pHead1->next; 
        } 
        else 
        { 
            q->next=pHead2; 
            pHead2=pHead2->next; 
        } 
        q=q->next; 
    } 
    if(pHead1) 
    { 
        while(pHead1) 
        { 
            q->next=pHead1; 
            q=q->next; 
            pHead1=pHead1->next; 
        } 
    } 
    if(pHead2) 
    { 
        while(pHead2) 
        { 
            q->next=pHead2; 
            q=q->next; 
            pHead2=pHead2->next; 
        } 
    } 
    return pMergedHead; 
} 

//非递归实现合并两个有序单链表(不额外开辟空间)
LinkNode* non_merge_list(LinkNode *pHead1, LinkNode *pHead2)
{
 if(pHead1==NULL)
  return pHead2;
 if(pHead2==NULL)
  return pHead1;
 if(pHead1==NULL && pHead2==NULL)
  return NULL;
 LinkNode *pMergedHead = NULL;
 LinkNode *q=NULL;
 if(pHead1->value<pHead2->value)
 {
  pMergedHead=pHead1;
  pHead1=pHead1->next;
 }
 else
 {
  pMergedHead=pHead2;
  pHead2=pHead2->next;
 } 
 q=pMergedHead;
 while(pHead1 && pHead2)
 {
  if(pHead1->value<pHead2->value)
  {
   q->next=pHead1;
   pHead1=pHead1->next;
  }
  else
  {
   q->next=pHead2;
   pHead2=pHead2->next;
  }
  q=q->next;
 }
 if(pHead1)
 {
  while(pHead1)
  {
   q->next=pHead1;
   q=q->next;
   pHead1=pHead1->next;
  }
 }
 if(pHead2)
 {
  while(pHead2)
  {
   q->next=pHead2;
   q=q->next;
   pHead2=pHead2->next;
  }
 }
 return pMergedHead;
}

2 输出单链表中的中间元素(若链表节点个数为偶数，则输出中间两个的任意一个)

思路：利用两个指针从头节点开始遍历，一个走一步，一个走两步，当一次走两步的指针走到链表末尾时，此时一次走一步的指针就指向链表的中间节点

代码如下：

LinkNode* print_mid_node(LinkNode *pHead) 
{ 
    LinkNode *pOne = pHead, *pTwo = pHead; 
    while(1)     
    { 
        pOne = pOne->next; 
        pTwo = pTwo->next->next; 
        if(pTwo==NULL || pTwo->next==NULL) 
            return pOne; 
    } 
} 

LinkNode* print_mid_node(LinkNode *pHead)
{
 LinkNode *pOne = pHead, *pTwo = pHead;
 while(1) 
 {
  pOne = pOne->next;
  pTwo = pTwo->next->next;
  if(pTwo==NULL || pTwo->next==NULL)
   return pOne;
 }
}

3 判断单恋表是否有环

思路与第二题一样，只是结束条件不一样，如果当一次走一步的指针等于一次走两步的指针时，则表示该链表有环

代码如下：

bool is_circle_list(LinkNode *pHead) 
{ 
    LinkNode *pOne = pHead, *pTwo = pHead; 
    while(1) 
    { 
        pOne = pOne->next; 
        pTwo = pTwo->next->next; 
        if(pOne == pTwo) 
            return true; 
        if(pTwo==NULL || pTwo->next==NULL) 
            return false; 
    } 
} 

bool is_circle_list(LinkNode *pHead)
{
 LinkNode *pOne = pHead, *pTwo = pHead;
 while(1)
 {
  pOne = pOne->next;
  pTwo = pTwo->next->next;
  if(pOne == pTwo)
   return true;
  if(pTwo==NULL || pTwo->next==NULL)
   return false;
 }
}

补充：软件开发 , C++ ,