20170401去哪儿笔试

来源：互联网发布：中华养生源码编辑：程序博客网时间：2024/04/27 23:19

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#include <iostream>#include <queue>using namespace std;//二叉树定义  struct BinaryTreeNode  {      int m_nValue;      BinaryTreeNode* m_pLeft;      BinaryTreeNode* m_pRight;      BinaryTreeNode(int x) : m_nValue(x), m_pLeft(NULL), m_pRight(NULL){}//构造函数  };  typedef BinaryTreeNode* binaryTreeNode;//根据前序和中序构建二叉树，核心函数  BinaryTreeNode* buildTreeAccordingPreAndIn(int preorder[], int inorder[],      int startPre,      int endPre,      int startIn,      int endIn){      if (startPre > endPre)          return NULL;      int nRootVal = preorder[startPre];      BinaryTreeNode* pRoot = new BinaryTreeNode(nRootVal);      int i, cnt = 0;      for (i = startIn; i <= endIn&&inorder[i] != nRootVal; i++, cnt++);      pRoot->m_pLeft = buildTreeAccordingPreAndIn(preorder, inorder, startPre + 1, startPre + cnt, startIn, i - 1);      pRoot->m_pRight = buildTreeAccordingPreAndIn(preorder, inorder, startPre + cnt + 1, endPre, i + 1, endIn);      return pRoot;  }  //根据前序和中序构建二叉树  BinaryTreeNode* buildTree(int preorder[],int preLength, int inorder[],int inLength) {      return buildTreeAccordingPreAndIn(preorder, inorder, 0, preLength - 1, 0, inLength - 1);  }  //按层遍历void leverTravel(BinaryTreeNode* pRoot)    {        queue<binaryTreeNode> q;        if(pRoot != NULL)        {            q.push(pRoot);        }        BinaryTreeNode* b;        while(!q.empty())        {            b = q.front();            cout<<b->m_nValue<<" ";            q.pop();            if(b->m_pLeft)            {                q.push(b->m_pLeft);            }            if(b->m_pRight)            {                q.push(b->m_pRight);            }        }    }int main()  {       int n;    cin>>n;      int preorder[n];     int inorder[n];    for(int i = 0; i < n; i++)    {        cin>>preorder[i];    }     for(int i = 0; i < n; i++)    {        cin>>inorder[i];    }    BinaryTreeNode* pRoot = buildTree(preorder,  n, inorder, n);      leverTravel(pRoot);     return 0;  }

牛客上做过原题吧，剑指offer的？构建二叉树，打印二叉树。

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#include <iostream>#include <math.h>using namespace std; int main(){    string str;    while(cin>>str)    {        int len = str.length();        long sum = 0;        long x;        for(int i = 0; i < len; i++)        {            x = str[i]-'a';            x *= pow(26, len - i - 1);            sum += x;        }        cout<<sum<<endl;    }     return 0;}

x如果定义为int 只能过80%。就是范围的问题

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#include <iostream>#include <unordered_set>#include <queue>using namespace std;int ladderLength(string start, string end, unordered_set<string> &dict) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        //BFS遍历找到的第一个匹配就是最短转换,空字符串是层与层之间的分隔标志        queue<string> Q;        Q.push(start); Q.push("");        int res = 1;        while(Q.empty() == false)        {            string str = Q.front();            Q.pop();            if(str != "")            {                int strLen = str.length();                for(int i = 0; i < strLen; i++)                {                    char tmp = str[i];                    for(char c = 'a'; c <= 'z'; c++)                    {                        if(c == tmp)continue;                        str[i] = c;                        if(str == end)return res+1;                        if(dict.find(str) != dict.end())                        {                            Q.push(str);                            dict.erase(str);                        }                    }                    str[i] = tmp;                }            }            else if(Q.empty() == false)            {//到达当前层的结尾，并且不是最后一层的结尾                res++;                Q.push("");            }        }        return 0;}int main(){    string start;    string end;    cin>>start;    cin>>end;    unordered_set<string> dict;    string str;    while(cin>>str)    {        dict.insert(str);    }    int n = ladderLength(start, end, dict);    cout<<n<<endl;}

Leetcode: Word Ladder原题

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