去哪儿网实习生笔试题

去哪儿网实习生笔试题目：.

第一题：

1.由前序遍历和中序遍历构造二叉树。

2.层次遍历二叉树并打印。

#include<iostream>#include<queue>using namespace std;struct TreeNode{    TreeNode*left;    TreeNode*right;    int val;    TreeNode(int x):left(NULL),right(NULL),val(x){}};TreeNode* builddfs(int a[],int prei,int prej,int b[],int ini,int inj){    if(ini>inj||prei>prej)    return NULL;    int target=a[prei];    int start=ini;    TreeNode*root=new TreeNode(a[prei]);    int len=0;    while(start<=inj&&b[start]!=target) {start++;len++;}    root->left=builddfs(a,prei+1,len+prei,b,ini,ini+len-1);    root->right=builddfs(a,len+prei+1,prej,b,start+1,inj);    return root;}void printbfs(TreeNode*root){    queue<TreeNode*> q;    q.push(root);    while(!q.empty())    {        TreeNode*front=q.front();        q.pop();        cout<<front->val<<" ";        if(front->left) q.push(front->left);        if(front->right) q.push(front->right);    }}int main(){    int n;    cin>>n;    int*pre=new int[n];    int*in=new int[n];    for(int i=0;i<n;i++) cin>>pre[i];    for(int i=0;i<n;i++) cin>>in[i];    TreeNode*root=builddfs(pre,0,n-1,in,0,n-1);    printbfs(root);    delete []pre;    delete []in;    return 0;}

第二题：

a,b,c,d,e,,,,,,x,y,z   的值为  0-25

26进制，由英文字符串求出10进制的值。

#include <iostream>#include <string>using namespace std;long long change(string str){    long long sum=0;    for(int i=0;i<str.size();i++)    {        sum=26*sum+str[i]-'a';    }    return sum;}int main(){    string str;    while(cin>>str)    {        cout<<change(str)<<endl;    }    return 0;}

第三题：编号127 Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

word ladder是一类比较难的题目。用DFS来做。每一步从单词里面拆掉一个字母替换，然后看看改变字母后的单词在不在词典里

#include <iostream>#include <unordered_set>#include <string>#include <queue>using namespace std;int ladderlength(string start,string end,unordered_set<string>&dict){    queue<string> path;    path.push(start);    int level=1;    int count=1;    dict.erase(start);    while(dict.size()>0&&!path.empty())    {        string curword=path.front();        path.pop();        count--;        for(int i=0;i<curword.size();i++)        {            string tmp=curword;            for(char j='a';j<='z';j++)            {                if(tmp[i]==j) continue;                tmp[i]=j;                if(tmp==end) return level+1;                if(dict.find(tmp)!=dict.end()) path.push(tmp);                dict.erase(tmp);            }        }        if(count==0)        {            count=path.size();            level++;        }    }    return 0;}int main(){        string str1;    string str2;    string tmp;    cin>>str1;    cin>>str2;    unordered_set<string> dict;    while(cin>>tmp)    {        dict.insert(tmp);    }    cout<< ladderlength(str1,str2,dict)<<endl;    return 0;}