面试准备_careercup_leetcode

1.3* Remove the duplicate charactersDesign an algorithm and write code to remove the duplicate characters in a string without using any additional buffer. NOTE: One or two additional variables are fine. An extra copy of the array is not. FOLLOW UPWrite the test cases for this method.Solution：use hash table of 256 elements.void removeDup(char* str){char hashtable[256];memset(hashtable, 0, 256*sizeof(char));int len = strlen(str);int idx = 0;for(int i=0; i<len; i++){if(hashtable[str[i]]==0){hashtable[str[i]] = 1;str[idx] = str[i];idx++;}}str[idx] = '\0';}1.5* Replace spaces in a stringWrite a method to replace all spaces in a string with ‘%20’.//以下代码适用于连续的空格算一个int replaceSpace(char* str, char*& newStr){int count = 0;int len = strlen(str);int charNum = 0;bool last = false;for(int i=0; i<len; i++){if(str[i] != ' '){charNum ++;if(last==true)last = false;}elseif(last==false&& str[i] ==' '){count++;last = true;}}int newLen = charNum + 3*count;newStr = newchar [newLen+1];newStr[newLen] = '\0';int newIdx = newLen-1;last = false;for(int idx=len-1; idx>=0; idx--){if(str[idx]!=' '){newStr[newIdx] = str[idx];newIdx--;if(last=true)last=false;}elseif(str[idx]==' '&& last==false){newStr[newIdx] = '0';newStr[newIdx-1] = '2';newStr[newIdx-2] = '%';newIdx -=3;last = true;}}return count;}1.6 矩阵旋转Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?1.8* Call substring to check rotationAssume you have a method isSubstring which checks if one word is a substring of another. Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubstring (i.e., “waterbottle” is a rotation of “erbottlewat”).解：将两个s1连接起来形成新的字符串s3, 看s2是否是s3的子串。

2.4 链表加法You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.EXAMPLEInput: (3 -> 1 -> 5), (5 -> 9 -> 2)Output: 8 -> 0 -> 82.5*链表中环的位置Given a circular linked list, implement an algorithm which returns node at the beginning of the loop. EXAMPLEinput: A -> B -> C -> D -> E -> C [the same C as earlier]output: CSolution:Step 1: fast, slow两个指针，fast每次走两步，slow每次走一步；假设当slow走到环的起点位置时走了K步，那么fast走了2K步，fast在环中K%L处，（即K为环的起始位置到头结点的距离，L为环中结点数）Step 2: 当他们相遇时，在环的起始位置前L-K%L 处；此时，将slow置为head，令fast和slow 同步走；Step 3: 当slow和fast再次相遇时，为环的起点处。

因为：设m=K%L, 则K=x·L+m，第二次相遇的位置为L-m+K=L-m+ x·L+m = nL, 是L的整数倍。

3.6* Sort a stackWrite a program to sort a stack in ascending order. You should not make any assumptions about how the stack is implemented. The following are the only functions that should be used to write this program: Push | Pop | Top | IsEmptyvoid sortStack(stack<int>& src, stack<int>& dst){if(src.empty())return;while(src.size()){int tmp = src.top();src.pop();while(!dst.empty() && tmp<dst.top()){int T=dst.top();dst.pop();src.push(T);}dst.push(tmp);}}void main(){stack<int> src, dst;for(int i=0; i<10; i++){int t = rand()%100;src.push(t);cout<<t<<" ";}sortStack(src, dst);cout<<endl;for(int i=0; i<10; i++){int t = dst.top();dst.pop();cout<<t<<" ";}system("pause");}4.2*图中两结点是否连通Given a directed graph, design an algorithm to find out whether there is a route between two nodes. Solution:template<typename T>bool Graph<T>::FindPath(VexNode<T>* startNode, VexNode<T>* endNode){stack<VexNode<T>*> travelStack;travelStack.push(startNode);for(int k=0; k<vexSize; k++){if(startNode == graph[k]){visited[k] = 1;break;}}while(!travelStack.empty()){VexNode<T>* topNode = travelStack.top();travelStack.pop();VisitFunc(topNode->data);ArcNode* arc = topNode->arcList;while(arc != NULL){if(graph[arc->vexIdx] == endNode)returntrue;if(visited[arc->vexIdx]==0){visited[arc->vexIdx] = 1;travelStack.push(graph[arc->vexIdx]);}arc = arc->pNext;}}returnfalse;}4.3* Create a binary tree with minimal heightGiven a sorted (increasing order) array, write an algorithm to create a binary tree with minimal height.BiTreeNode* ConstructLowest(int data[], int startIdx, int endIdx){if(startIdx > endIdx)return NULL;int mid = (endIdx + startIdx)/2;BiTreeNode* node = new BiTreeNode;node->m_value = data[mid];node->m_pLeft = node->m_pRight = NULL;if(mid-1>=startIdx)node->m_pLeft = ConstructLowest(data, startIdx, mid-1);if(mid+1<=endIdx)node->m_pRight = ConstructLowest(data, mid+1, endIdx);return node;}4.4*树中每层的结点Given a binary search tree, design an algorithm which creates a linked list of all the nodes at each depth (i.e., if you have a tree with depth D, you’ll have D linked lists).void levelNodeList(BiTreeNode* tree, vector<list<BiTreeNode*>>& result){if(tree==NULL)return;list<BiTreeNode*> alevel, prelevel;alevel.push_back(tree);result.push_back(alevel);prelevel = alevel;int prelevelSize = 1;while(prelevelSize){list<BiTreeNode*> newlevel;list<BiTreeNode*>::iterator it;int n = 0;for(it = prelevel.begin(); it != prelevel.end(); it++){if((*it)->m_pLeft){newlevel.push_back((*it)->m_pLeft);n++;}if((*it)->m_pRight){newlevel.push_back((*it)->m_pRight);n++;}}prelevelSize = n;if(newlevel.size()){result.push_back(newlevel);prelevel = newlevel;}}}4.5* Next node in in-order successorWrite an algorithm to find the ‘next’ node (i.e., in-order successor) of a given node in a binary search tree where each node has a link to its parent.struct BiSearchNode{int m_value;BiSearchNode* m_pLeft;BiSearchNode* m_pRight;BiSearchNode* m_pParent;};BiSearchNode* getSuccessor(BiSearchNode* tree, BiSearchNode* anode){if(!tree || !anode)return NULL;BiSearchNode* parent = anode->m_pParent;if(parent && anode == parent->m_pLeft) //该结点是其父节点的左子树{return parent;}if(parent==NULL || (parent && anode == parent->m_pRight)) //该结点是其父节点的右子树{if(anode->m_pRight) //该结点有右子树，后继是右子树的最左结点{BiSearchNode* tmp = anode->m_pRight;while(tmp->m_pLeft)tmp = tmp->m_pLeft;return tmp;}else//该结点无右子树，延父节点向上直到某结点是其父节点的左子结点{BiSearchNode* tmp = parent;parent = parent->m_pParent;while(parent && tmp == parent->m_pRight){tmp = parent;parent = parent->m_pParent;}if(!parent)return NULL;elseif(tmp==parent->m_pLeft)return parent;}}}4.6*最低公共祖先，不用辅助空间Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree. 思路：所谓最低公共祖先，设r为p、q的最低公共祖先，则p、q中一个节点在r的左边，一个节点在r的右边（暂时不考虑p是q的祖先的情况），则只需要统计一下一颗子树是否包含p，q，如果p、q都在一个节点的左子树，在继续判断该节点的左孩子，直到p、q分别位于节点的左右两边。