程序员面试题精选100题

1、程序员面试题精选100题折叠 前言随着高校的持续扩张，每年应届毕业生的数目都在不断增长，伴随而来的是应届毕业生的就业压力也越来越大。 在这样的背景下，就业变成一个买方市场的趋势越来越明显。为了找到一个称心的工作，绝大多数应届毕业生都必须反复经历简历筛选、电话面试、笔试、面试等环节。在这些环节中，面试无疑起到最为重要的作用，因为通过面试公司能够最直观的了解学生的能力。为了有效地准备面试，面经这个新兴概念应运而生。笔者在当初找工作阶段也从面经中获益匪浅并最终找到满意的工作。为了方便后来者，笔者花费大量时间收集并整理散落在茫茫网络中的面经。不同行业的面经全然不同，笔者从自身专业出发，着重关注程序员面试的面经，并从精选出若干具有代表性的技术类的面试题展开讨论，希望能给读者带来一些启发。由于笔者水平有限，给各面试题提供的思路和代码难免会有错误，还请读者批评指正。另外，热忱欢迎读者能够提供更多、更好的面试题，本人将感激不尽。(01)把二元查找树转变成排序的双向链表折叠 题目：输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。要求不能创建任何新的结点，只调整指针的指向。比如将二元查找树 1

2、0 / 6 14 / / 4 8 12 16转换成双向链表4=6=8=10=12=14=16。分析：本题是微软的面试题。很多与树相关的题目都是用递归的思路来解决，本题也不例外。下面我们用两种不同的递归思路来分析。思路一：当我们到达某一结点准备调整以该结点为根结点的子树时，先调整其左子树将左子树转换成一个排好序的左子链表，再调整其右子树转换右子链表。最近链接左子链表的最右结点（左子树的最大结点）、当前结点和右子链表的最左结点（右子树的最小结点）。从树的根结点开始递归调整所有结点。思路二：我们可以中序遍历整棵树。按照这个方式遍历树，比较小的结点先访问。如果我们每访问一个结点，假设之前访问过的结点已经调整成一个排序双向链表，我们再把调整当前结点的指针将其链接到链表的末尾。当所有结点都访问过之后，整棵树也就转换成一个排序双向链表了。参考代码：首先我们定义二元查找树结点的数据结构如下： struct BSTreeNode / a node in the binary search tree int m_nValue; / value of nodeBSTreeNode *m_pLeft; / l

3、eft child of nodeBSTreeNode *m_pRight; / right child of node ;思路一对应的代码：/ Covert a sub binary-search-tree into a sorted double-linked list/ Input: pNode - the head of the sub tree/ asRight - whether pNode is the right child of its parent/ Output: if asRight is true, return the least node in the sub-tree/ else return the greatest node in the sub-tree/BSTreeNode* ConvertNode(BSTreeNode* pNode, bool asRight)if(!pNode)return NULL;BSTreeNode *pLeft = NULL;BSTreeNode *pRight = NULL;/ Convert the left s

4、ub-treeif(pNode-m_pLeft)pLeft = ConvertNode(pNode-m_pLeft, false);/ Connect the greatest node in the left sub-tree to the current nodeif(pLeft)pLeft-m_pRight = pNode;pNode-m_pLeft = pLeft;/ Convert the right sub-treeif(pNode-m_pRight)pRight = ConvertNode(pNode-m_pRight, true);/ Connect the least node in the right sub-tree to the current nodeif(pRight)pNode-m_pRight = pRight;pRight-m_pLeft = pNode;BSTreeNode *pTemp = pNode;/ If the current node is the right child of its parent,/ return the least

5、node in the tree whose root is the current nodeif(asRight)while(pTemp-m_pLeft)pTemp = pTemp-m_pLeft;/ If the current node is the left child of its parent,/ return the greatest node in the tree whose root is the current nodeelsewhile(pTemp-m_pRight)pTemp = pTemp-m_pRight;return pTemp;/ Covert a binary search tree into a sorted double-linked list/ Input: the head of tree/ Output: the head of sorted double-linked list/BSTreeNode* Convert(BSTreeNode* pHeadOfTree)/ As we want to return the head of th

6、e sorted double-linked list,/ we set the second parameter to be truereturn ConvertNode(pHeadOfTree, true);思路二对应的代码：/ Covert a sub binary-search-tree into a sorted double-linked list/ Input: pNode - the head of the sub tree/ pLastNodeInList - the tail of the double-linked list/void ConvertNode(BSTreeNode* pNode, BSTreeNode*& pLastNodeInList)if(pNode = NULL)return;BSTreeNode *pCurrent = pNode;/ Convert the left sub-treeif (pCurrent-m_pLeft != NULL)ConvertNode(pCurrent-m_pLeft, pLastNodeInList);/ P

7、ut the current node into the double-linked listpCurrent-m_pLeft = pLastNodeInList;if(pLastNodeInList != NULL)pLastNodeInList-m_pRight = pCurrent;pLastNodeInList = pCurrent;/ Convert the right sub-treeif (pCurrent-m_pRight != NULL)ConvertNode(pCurrent-m_pRight, pLastNodeInList);/ Covert a binary search tree into a sorted double-linked list/ Input: pHeadOfTree - the head of tree/ Output: the head of sorted double-linked list/BSTreeNode* Convert_Solution1(BSTreeNode* pHeadOfTree)BSTreeNode *pLastNodeInList = NULL;ConvertNode(pHeadOfTree, pLastNodeInList);/ Get the head of the double-linked listBSTreeNode *pHeadOfList = pLastNodeInList;while(pHeadOfList & pHeadOfList-m_pLeft)pHeadOfList

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