找工过程中碰到的笔试面试题整理（2）

继续我的找工笔试面试题整理。

1.获取字符串最长数字串

这个是baidu的一道笔试题，虽然简单，但是要当场写对还是要仔细点。

 

代码

   
     

    
      
    
      int
    
       getlongestdigital(
    
      const
    
       
    
      char
    
      *
    
       inputstr,
    
      char
    
      *
    
       outputstr)
{
 
    
      int
    
       start,length,longeststart,longest,offset;
 start 
    
      =
    
       length 
    
      =
    
       longest 
    
      =
    
       longeststart 
    
      =
    
       offset 
    
      =
    
       
    
      0
    
      ;
 
    
      while
    
      (
    
      *
    
      inputstr
    
      !=
    
      '
    
      \0
    
      '
    
      )
 {
 
    
      if
    
      (
    
      *
    
      inputstr
    
      >=
    
      '
    
      0
    
      '
    
      &&*
    
      inputstr
    
      <=
    
      '
    
      9
    
      '
    
      )
 {
 
    
      if
    
      (length
    
      ==
    
      0
    
      )
 start 
    
      =
    
       offset;
 length
    
      ++
    
      ;
 }
 
    
      else
    
      
 {
 
    
      if
    
      (length
    
      >
    
      longest)
 {
 longest 
    
      =
    
       length;
 longeststart 
    
      =
    
       start;
 }
 length 
    
      =
    
      0
    
      ;
 }
 offset
    
      ++
    
      ;
 inputstr
    
      ++
    
      ;
 }
 
    
      if
    
      (length
    
      >
    
      longest)
 {
 longest 
    
      =
    
       length;
 longeststart 
    
      =
    
       start;
 }
 inputstr 
    
      -=
    
       offset;
 
    
      for
    
      (
    
      int
    
       i
    
      =
    
      0
    
      ;i
    
      <
    
      longest;i
    
      ++
    
      )
 {
 outputstr[i] 
    
      =
    
       inputstr[i
    
      +
    
      longeststart];
 }
 outputstr[longest] 
    
      =
    
       
    
      '
    
      \0
    
      '
    
      ;
 
    
      return
    
       longest;
}
   
     

 

2.完全二叉树的节点添加

这个也是常出现的题目，完全二叉树是笔试面试经常碰到的，最好要熟悉的它的各种操作。

 

 

代码

    
      

     
       
     
       typedef 
     
       struct
     
        TreeNode
{
 
     
       int
     
        data;
 
     
       struct
     
        TreeNode
     
       *
     
        left;
 
     
       struct
     
        TreeNode
     
       *
     
        right;
}TreeNode;

     
       void
     
        insertnode(TreeNode
     
       *
     
        root,TreeNode
     
       *
     
        newnode)
{
 queue
     
       <
     
       TreeNode
     
       *>
     
        nonvisit;
 nonvisit.push(root);
 
     
       while
     
       (
     
       !
     
       nonvisit.empty())
 {
 TreeNode
     
       *
     
        pvisit 
     
       =
     
        nonvisit.front();
 nonvisit.pop();
 
     
       if
     
       (pvisit
     
       ->
     
       left
     
       ==
     
       NULL)
 {
 pvisit
     
       ->
     
       left 
     
       =
     
        newnode;
 
     
       return
     
       ;
 }
 nonvisit.push(pvisit
     
       ->
     
       left);
 
     
       if
     
       (pvisit
     
       ->
     
       right 
     
       ==
     
        NULL)
 {
 pvisit
     
       ->
     
       right 
     
       =
     
        newnode;
 
     
       return
     
       ;
 }
 nonvisit.push(pvisit
     
       ->
     
       right);
 }
}
    
      

 

3.链表反序

 

链表是又一个重点，熟悉一下它的各种操作吧

 

代码

    
      

     
       
     
       void
     
        ReverseList(LinkNode 
     
       **
     
       root)
{
 LinkNode 
     
       *
     
       temp1 
     
       =
     
        (
     
       *
     
       root)
     
       ->
     
       next;
 
     
       if
     
       (temp1
     
       ==
     
       NULL)
 
     
       return
     
       ;
 LinkNode 
     
       *
     
       temp2;
 (
     
       *
     
       root)
     
       ->
     
       next 
     
       =
     
        NULL;
 temp2 
     
       =
     
        temp1
     
       ->
     
       next;
 temp1
     
       ->
     
       next 
     
       =
     
        
     
       *
     
       root;
 
     
       *
     
       root 
     
       =
     
        temp1;
 
     
       while
     
       (temp2
     
       !=
     
       NULL)
 {
 temp1 
     
       =
     
        temp2;
 temp2 
     
       =
     
        temp1
     
       ->
     
       next;
 temp1
     
       ->
     
       next 
     
       =
     
        
     
       *
     
       root;
 
     
       *
     
       root 
     
       =
     
        temp1;
 }
}