剑指Offer-第2章 面试需要的基础知识

面试题1: 赋值运算符函数

一般的写法是如下形式:

 

CMyString& operator=(const CMyString& str){            if(this!=&str){                delete[] m_pData;                int sz=strlen(str.m_pData);                m_pData=new char[sz+1];                strcpy(m_pData,str.m_pData);            }            return *this;        }

CMyString& operator=(const CMyString& str){            if(this!=&str){                CMyString tmp_str(str);                char *p=m_pData;                m_pData=tmp_str.m_pData;                tmp_str.m_pData=p;            }            return *this;        }

完整的测试实例如下:

class CMyString{    public:        CMyString(char *pData=NULL){            if(pData==NULL){                m_pData=new char[1];                m_pData[0]='\0';            }            else{                int sz=strlen(pData);                m_pData=new char[sz+1];                strcpy(m_pData,pData);            }        }        CMyString(const CMyString& str){            int sz=strlen(str.m_pData);            m_pData=new char[sz+1];            strcpy(m_pData,str.m_pData);        }        ~CMyString(){            delete[] m_pData;        }        //经典写法，适用于初级程序员        /*        CMyString& operator=(const CMyString& str){            if(this!=&str){                delete[] m_pData;                int sz=strlen(str.m_pData);                m_pData=new char[sz+1];                strcpy(m_pData,str.m_pData);            }            return *this;        }         */        //考虑异常安全性，高级写法        CMyString& operator=(const CMyString& str){            if(this!=&str){                CMyString tmp_str(str);                char *p=m_pData;                m_pData=tmp_str.m_pData;                tmp_str.m_pData=p;            }            return *this;        }        void print(ostream& os)const {            os<<m_pData;        }    private:        char* m_pData;};       ostream& operator<<(ostream& os,const CMyString& str){           str.print(os);            return os;        }void Test1(){    printf("Test1 begins:\n");    char* text = "Hello world";    CMyString str1(text);    CMyString str2;    str2 = str1;    printf("The expected result is: %s.\n", text);    printf("The actual result is: ");    cout<<str2<<endl;}// 赋值给自己void Test2(){    printf("Test2 begins:\n");    char* text = "Hello world";    CMyString str1(text);    str1 = str1;    printf("The expected result is: %s.\n", text);    printf("The actual result is: ");    cout<<str1<<endl;}// 连续赋值void Test3(){    printf("Test3 begins:\n");    char* text = "Hello world";    CMyString str1(text);    CMyString str2, str3;    str3 = str2 = str1;    printf("The expected result is: %s.\n", text);    printf("The actual result is: ");    cout<<str2<<endl;    printf("The expected result is: %s.\n", text);    printf("The actual result is: ");    cout<<str3<<endl;}int main(){    Test1();    Test2();    Test3();    return 0;}

面试题2: 实现singleton模式

//这是基本的形式class singleton_{    public:        static singleton_* get_instance(){            if(_instance==0)                _instance=new singleton_;            return _instance;        }    private:        singleton_(){}        ~singleton_(){}        static singleton_* _instance;};singleton_* singleton_::_instance=0;//下面的方法可以应用在多线程中，主要是利用锁的帮助class locker{    public:        locker(){            pthread_mutex_init(&mutex,NULL);        }        ~locker(){            pthread_mutex_destroy(&mutex);        }        void lock(){            pthread_mutex_lock(&mutex);        }        void unlock(){            pthread_mutex_unlock(&mutex);        }    private:        pthread_mutex_t mutex;};class singleton{    public:        static singleton* get_instance();    private:        singleton(){}        ~singleton(){}    private:        static singleton* _instance;        static locker llock;};singleton* singleton::_instance=0;locker singleton::llock;singleton* singleton::get_instance(){    if(_instance==0){        llock.lock();        printf("locked\n");        if(_instance==0)            _instance=new singleton;        llock.unlock();        printf("unlocked\n");    }    return _instance;}int main(){    int pid;    for(int i=0;i<2;i++){        pid=fork();        if(pid<0)            printf("error\n");        else if(pid==0){            singleton* slt=singleton::get_instance();            printf("child:%x\n",slt);        }        else{            singleton* slt=singleton::get_instance();            printf("parent:%x\n",slt);        }    }}

参考: http://blog.csdn.net/joanlynnlove/article/details/7462254

面试题3: 二维数组中的查找

代码如下:

bool FindInPartiallySortedMatrix(int *matrix,int row,int col,int number){    if(matrix==0||row<0||col<0)        return false;    int i=0,j=col-1;    while(i<row&&j>=0){        int t=matrix[i*col+j];        if(t==number)            return true;        else if(t>number)            --j;        else            i++;    }    return false;}// ====================测试代码====================void Test(char* testName, int* matrix, int rows, int columns, int number, bool expected){    if(testName != NULL)        printf("%s begins: ", testName);    bool result = FindInPartiallySortedMatrix(matrix, rows, columns, number);    if(result == expected)        printf("Passed.\n");    else        printf("Failed.\n");}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数在数组中void Test1(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test1", (int*)matrix, 4, 4, 7, true);}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数不在数组中void Test2(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test2", (int*)matrix, 4, 4, 5, false);}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数是数组中最小的数字void Test3(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test3", (int*)matrix, 4, 4, 1, true);}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数是数组中最大的数字void Test4(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test4", (int*)matrix, 4, 4, 15, true);}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数比数组中最小的数字还小void Test5(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test5", (int*)matrix, 4, 4, 0, false);}//  1   2   8   9//  2   4   9   12//  4   7   10  13//  6   8   11  15// 要查找的数比数组中最大的数字还大void Test6(){    int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};    Test("Test6", (int*)matrix, 4, 4, 16, false);}// 鲁棒性测试，输入空指针void Test7(){    Test("Test7", NULL, 0, 0, 16, false);}int main(int argc, char* argv[]){    Test1();    Test2();    Test3();    Test4();    Test5();    Test6();    Test7();    return 0;}

面试题4: 替换空格

这儿的关键问题是替换后字符串的长度变大了, 所以主要的问题是如何在低复杂度情况下完成复制.

代码如下:

void replace_blank(char *str,int capacity){    if(str==0||capacity<0)        return ;    int blk_cnt=0;    int len=0;    while(str[len]!='\0') {        if(str[len]==' ')            blk_cnt++;        len++;    }    if(len+1+blk_cnt*2>capacity)        return ;    char *p1,*p2;    p1=str+len+1;    p2=str+len+1+blk_cnt*2;    while(p1>=str&&p2>=p1){        if(*p1==' '){            p1--;            *p2--='0';            *p2--='2';            *p2--='%';        }        else            *p2--=*p1--;    }}void Test(char* testName, char string[], int length, char expected[]){    if(testName != NULL)        printf("%s begins: ", testName);    replace_blank(string, length);    if(expected == NULL && string == NULL)        printf("passed.\n");    else if(expected == NULL && string != NULL)        printf("failed.\n");    else if(strcmp(string, expected) == 0)        printf("passed.\n");    else        printf("failed.\n");}// 空格在句子中间void Test1(){    const int length = 100;    char string[length] = "hello world";    Test("Test1", string, length, "hello%20world");}// 空格在句子开头void Test2(){    const int length = 100;    char string[length] = " helloworld";    Test("Test2", string, length, "%20helloworld");}// 空格在句子末尾void Test3(){    const int length = 100;    char string[length] = "helloworld ";    Test("Test3", string, length, "helloworld%20");}// 连续有两个空格void Test4(){    const int length = 100;    char string[length] = "hello  world";    Test("Test4", string, length, "hello%20%20world");}// 传入NULLvoid Test5(){    Test("Test5", NULL, 0, NULL);}// 传入内容为空的字符串void Test6(){    const int length = 100;    char string[length] = "";    Test("Test6", string, length, "");}//传入内容为一个空格的字符串void Test7(){    const int length = 100;    char string[length] = " ";    Test("Test7", string, length, "%20");}// 传入的字符串没有空格void Test8(){    const int length = 100;    char string[length] = "helloworld";    Test("Test8", string, length, "helloworld");}// 传入的字符串全是空格void Test9(){    const int length = 100;    char string[length] = "   ";    Test("Test9", string, length, "%20%20%20");}int main(int argc, char* argv[]){    Test1();    Test2();    Test3();    Test4();    Test5();    Test6();    Test7();    Test8();    Test9();    return 0;}

面试题5: 从尾到头打印链表

本题目中需要用到链表的定义:

struct ListNode{    int m_nValue;    ListNode* m_pNext;};ListNode* CreateListNode(int value){    ListNode* pNode = new ListNode();    pNode->m_nValue = value;    pNode->m_pNext = NULL;    return pNode;}void ConnectListNodes(ListNode* pCurrent, ListNode* pNext){    if(pCurrent == NULL)    {        printf("Error to connect two nodes.\n");        exit(1);    }    pCurrent->m_pNext = pNext;}void PrintListNode(ListNode* pNode){     if(pNode == NULL)    {        printf("The node is NULL\n");    }    else    {        printf("The key in node is %d.\n", pNode->m_nValue);    }}void PrintList(ListNode* pHead){    printf("PrintList starts.\n");        ListNode* pNode = pHead;    while(pNode != NULL)    {        printf("%d\t", pNode->m_nValue);        pNode = pNode->m_pNext;    }    printf("\nPrintList ends.\n");}void DestroyList(ListNode* pHead){    ListNode* pNode = pHead;    while(pNode != NULL)    {        pHead = pHead->m_pNext;        delete pNode;        pNode = pHead;    }}void AddToTail(ListNode** pHead, int value){    ListNode* pNew = new ListNode();    pNew->m_nValue = value;    pNew->m_pNext = NULL;    if(*pHead == NULL)    {        *pHead = pNew;    }    else    {        ListNode* pNode = *pHead;        while(pNode->m_pNext != NULL)            pNode = pNode->m_pNext;        pNode->m_pNext = pNew;    }}void RemoveNode(ListNode** pHead, int value){    if(pHead == NULL || *pHead == NULL)        return;    ListNode* pToBeDeleted = NULL;    if((*pHead)->m_nValue == value)    {        pToBeDeleted = *pHead;        *pHead = (*pHead)->m_pNext;    }    else    {        ListNode* pNode = *pHead;        while(pNode->m_pNext != NULL && pNode->m_pNext->m_nValue != value)            pNode = pNode->m_pNext;        if(pNode->m_pNext != NULL && pNode->m_pNext->m_nValue == value)        {            pToBeDeleted = pNode->m_pNext;            pNode->m_pNext = pNode->m_pNext->m_pNext;        }    }    if(pToBeDeleted != NULL)    {        delete pToBeDeleted;        pToBeDeleted = NULL;    }}

本题的代码如下:

void PrintListReversingly_Iteratively(ListNode* pHead){    if(pHead==0)        return ;    stack<ListNode*> st;    ListNode* p=pHead;    while(p!=0){        st.push(p);        p=p->m_pNext;    }    while(!st.empty()){        p=st.top();st.pop();        cout<<p->m_nValue<<' ';    }    cout<<endl;}void PrintListReversingly_Recursively(ListNode* pHead){    if(pHead==0)        return;    PrintListReversingly_Recursively(pHead->m_pNext);    cout<<pHead->m_nValue<<' ';}void Test(ListNode* pHead){    PrintList(pHead);    PrintListReversingly_Iteratively(pHead);    printf("\n");    PrintListReversingly_Recursively(pHead);}// 1->2->3->4->5void Test1(){    printf("\nTest1 begins.\n");    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode4);    ConnectListNodes(pNode4, pNode5);    Test(pNode1);    DestroyList(pNode1);}// 只有一个结点的链表: 1void Test2(){    printf("\nTest2 begins.\n");    ListNode* pNode1 = CreateListNode(1);    Test(pNode1);    DestroyList(pNode1);}// 空链表void Test3(){    printf("\nTest3 begins.\n");    Test(NULL);}int main(int argc, char* argv[]){    Test1();    Test2();    Test3();    return 0;}

面试题6: 重建二叉树

二叉树相关的定义:

struct BinaryTreeNode {    int                    m_nValue;     BinaryTreeNode*        m_pLeft;      BinaryTreeNode*        m_pRight; };BinaryTreeNode* CreateBinaryTreeNode(int value){    BinaryTreeNode* pNode = new BinaryTreeNode();    pNode->m_nValue = value;    pNode->m_pLeft = NULL;    pNode->m_pRight = NULL;    return pNode;}void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight){    if(pParent != NULL)    {        pParent->m_pLeft = pLeft;        pParent->m_pRight = pRight;    }}void PrintTreeNode(BinaryTreeNode* pNode){    if(pNode != NULL)    {        printf("value of this node is: %d\n", pNode->m_nValue);        if(pNode->m_pLeft != NULL)            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);        else            printf("left child is null.\n");        if(pNode->m_pRight != NULL)            printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);        else            printf("right child is null.\n");    }    else    {        printf("this node is null.\n");    }    printf("\n");}void PrintTree(BinaryTreeNode* pRoot){    PrintTreeNode(pRoot);    if(pRoot != NULL)    {        if(pRoot->m_pLeft != NULL)            PrintTree(pRoot->m_pLeft);        if(pRoot->m_pRight != NULL)            PrintTree(pRoot->m_pRight);    }}void DestroyTree(BinaryTreeNode* pRoot){    if(pRoot != NULL)    {        BinaryTreeNode* pLeft = pRoot->m_pLeft;        BinaryTreeNode* pRight = pRoot->m_pRight;        delete pRoot;        pRoot = NULL;        DestroyTree(pLeft);        DestroyTree(pRight);    }}

本题的代码如下:

BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder);BinaryTreeNode* Construct(int* preorder, int* inorder, int length){    if(preorder==0||inorder==0||length<0)        return 0;    return ConstructCore(preorder,preorder+length-1,inorder,inorder+length-1);}BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder){    //判断输入是否合法    if(startPreorder>endPreorder||startInorder>endInorder)        return 0;    int rootValue=startPreorder[0];    BinaryTreeNode* root=new BinaryTreeNode;    root->m_nValue=rootValue;    root->m_pLeft=root->m_pRight=0;    if(startPreorder==endPreorder){        if(startInorder==endInorder&&*startPreorder==*startInorder)        return root;        else            throw exception();    }    int *rootInorder=startInorder;    while(rootInorder<=endInorder&&*rootInorder!=rootValue)        ++rootInorder;    if(rootInorder>endInorder||rootInorder==endInorder&&*rootInorder!=rootValue)            throw std::exception();    int len=rootInorder-startInorder;    //对len的检查和开头的判断检查的效果是一样的。//    if(len>0)    root->m_pLeft=ConstructCore(startPreorder+1,startPreorder+len,startInorder,rootInorder-1);//    if(len<endPreorder-startPreorder)    root->m_pRight=ConstructCore(startPreorder+len+1,endPreorder,rootInorder+1,endInorder);    return root;}// ====================测试代码====================void Test(char* testName, int* preorder, int* inorder, int length){    if(testName != NULL)        printf("%s begins:\n", testName);    printf("The preorder sequence is: ");    for(int i = 0; i < length; ++ i)        printf("%d ", preorder[i]);    printf("\n");    printf("The inorder sequence is: ");    for(int i = 0; i < length; ++ i)        printf("%d ", inorder[i]);    printf("\n");    try    {        BinaryTreeNode* root = Construct(preorder, inorder, length);        PrintTree(root);        DestroyTree(root);    }    catch(std::exception& exception)    {        printf("Invalid Input.\n");    }}// 普通二叉树//              1//           /     \//          2       3  //         /       / \//        4       5   6//         \         ///          7       8void Test1(){    const int length = 8;    int preorder[length] = {1, 2, 4, 7, 3, 5, 6, 8};    int inorder[length] = {4, 7, 2, 1, 5, 3, 8, 6};    Test("Test1", preorder, inorder, length);}// 所有结点都没有右子结点//            1//           / //          2   //         / //        3 //       ///      4//     ///    5void Test2(){    const int length = 5;    int preorder[length] = {1, 2, 3, 4, 5};    int inorder[length] = {5, 4, 3, 2, 1};    Test("Test2", preorder, inorder, length);}// 所有结点都没有左子结点//            1//             \ //              2   //               \ //                3 //                 \//                  4//                   \//                    5void Test3(){    const int length = 5;    int preorder[length] = {1, 2, 3, 4, 5};    int inorder[length] = {1, 2, 3, 4, 5};    Test("Test3", preorder, inorder, length);}// 树中只有一个结点void Test4(){    const int length = 1;    int preorder[length] = {1};    int inorder[length] = {1};    Test("Test4", preorder, inorder, length);}// 完全二叉树//              1//           /     \//          2       3  //         / \     / \//        4   5   6   7void Test5(){    const int length = 7;    int preorder[length] = {1, 2, 4, 5, 3, 6, 7};    int inorder[length] = {4, 2, 5, 1, 6, 3, 7};    Test("Test5", preorder, inorder, length);}// 输入空指针void Test6(){    Test("Test6", NULL, NULL, 0);}// 输入的两个序列不匹配void Test7(){    const int length = 7;    int preorder[length] = {1, 2, 4, 5, 3, 6, 7};    int inorder[length] = {4, 2, 8, 1, 6, 3, 7};    Test("Test7: for unmatched input", preorder, inorder, length);}int main(int argc, char* argv[]){    Test1();    Test2();    Test3();    Test4();    Test5();    Test6();    Test7();    return 0;}

面试题7: 用两个栈实现队列操作

template <typename T> class CQueue{public:    CQueue(void);    ~CQueue(void);        // 在队列末尾添加一个结点    void appendTail(const T& node);    // 删除队列的头结点    T deleteHead();private:    stack<T> stack1;    stack<T> stack2;};template <typename T> CQueue<T>::CQueue(void){}template <typename T> CQueue<T>::~CQueue(void){}template<typename T> void CQueue<T>::appendTail(const T& element){    stack1.push(element);} template<typename T> T CQueue<T>::deleteHead(){    if(stack2.size()<= 0)    {        while(stack1.size()>0)        {            T& data = stack1.top();            stack1.pop();            stack2.push(data);        }    }    if(stack2.size() == 0)        throw exception();    T head = stack2.top();    stack2.pop();    return head;}void Test(char actual, char expected){    if(actual == expected)        printf("Test passed.\n");    else        printf("Test failed.\n");}int main(int argc, char* argv[]){    CQueue<char> queue;    queue.appendTail('a');    queue.appendTail('b');    queue.appendTail('c');    char head = queue.deleteHead();    Test(head, 'a');    head = queue.deleteHead();    Test(head, 'b');    queue.appendTail('d');    head = queue.deleteHead();    Test(head, 'c');    queue.appendTail('e');    head = queue.deleteHead();    Test(head, 'd');    head = queue.deleteHead();    Test(head, 'e');    return 0;}

面试题8: 旋转数组的最小数字

很多题目会用到快排的思想, 快排中最核心的地方就是利用一个数将数组分为两部分, 实现方法如下:

方法1: 从两边往中间遍历

int Partition(int *a,int low,int high){    if(a== NULL || low>high)        throw new std::exception();    int pivot=a[low];    int i=low,j=high+1;    while(1){        do            i++;        while(a[i]<pivot);        do            j--;        while(a[j]>pivot);        if(i>=j)            break;        swap(a[i],a[j]);    }    swap(a[low],a[j]);    return j;}

方法2: 从一边往另一边遍历,比较巧

int RandomInRange(int minl,int maxl){    return rand()%(maxl-minl+1)+minl;}int Partition_(int data[],int start, int end){    if(data == NULL || start >end )        throw new std::exception();    int index = RandomInRange(start, end);    swap(data[index], data[end]);    int small = start - 1;    for(index = start; index < end; ++ index)    {        if(data[index] < data[end])        {            ++ small;            if(small != index)                swap(data[index], data[small]);        }    }    ++ small;    swap(data[small], data[end]);    return small;}

代码如下:

int MinInOrder(int* numbers, int index1, int index2);int Min(int *arr,int len){    if(arr==0||len<=0)        throw exception();    int i=0,j=len-1;    int mid=i;    while(arr[i]>=arr[j]){        //如果前后指针指向相邻的两个数，那么后面的那个数就是我们所要查找的。        if(j-i==1){            mid=j;            break;        }        mid=(i+j)/2;        //如果三个数相同，只能采取线性查找了。        if(arr[i]==arr[j]&&                arr[mid]==arr[i])            return MinInOrder(arr,i,j);        //缩小查找范围        if(arr[mid]>=arr[i])            i=mid;       else if(arr[mid]<=arr[j])            j=mid;    }    return arr[mid];}//线性查找int MinInOrder(int *a,int l,int r){    int res=a[l];    for(int i=l;i<r;i++)        if(res>a[i])            res=a[i];    return res;}// ====================测试代码====================void Test(int* numbers, int length, int expected){    int result = 0;    try    {        result = Min(numbers, length);        for(int i = 0; i < length; ++i)            printf("%d ", numbers[i]);        if(result == expected)            printf("\tpassed\n");        else            printf("\tfailed\n");    }    catch (...)    {        if(numbers == NULL)            printf("Test passed.\n");        else            printf("Test failed.\n");    }}int main(int argc, char* argv[]){    // 典型输入，单调升序的数组的一个旋转    int array1[] = {3, 4, 5, 1, 2};    Test(array1, sizeof(array1) / sizeof(int), 1);    // 有重复数字，并且重复的数字刚好的最小的数字    int array2[] = {3, 4, 5, 1, 1, 2};    Test(array2, sizeof(array2) / sizeof(int), 1);    // 有重复数字，但重复的数字不是第一个数字和最后一个数字    int array3[] = {3, 4, 5, 1, 2, 2};    Test(array3, sizeof(array3) / sizeof(int), 1);    // 有重复的数字，并且重复的数字刚好是第一个数字和最后一个数字    int array4[] = {1, 0, 1, 1, 1};    Test(array4, sizeof(array4) / sizeof(int), 0);    // 单调升序数组，旋转0个元素，也就是单调升序数组本身    int array5[] = {1, 2, 3, 4, 5};    Test(array5, sizeof(array5) / sizeof(int), 1);    // 数组中只有一个数字    int array6[] = {2};    Test(array6, sizeof(array6) / sizeof(int), 2);    // 输入NULL    Test(NULL, 0, 0);    return 0;}

面试题9: 斐波那契数列

这儿总结一下所有求解斐波那契数列的方法:

方法1:

经典的递归方式.

方法2:

迭代的方式

方法3:

比较巧的方法, 根据递推表达式得到一个矩阵的表示形式:

三种方法的代码如下:

int Fibonacci1(int n){    if(n==0)        return 0;    if(n==1)        return 1;    return Fibonacci1(n-1)+Fibonacci1(n-2);}int Fibonacci2(int n){    if(n==0)        return 0;    if(n==1)        return 1;    int fn,fn_1,fn_2;    fn_1=1,fn_2=0;    for(int i=2;i<=n;i++){        fn=fn_1+fn_2;        fn_2=fn_1;        fn_1=fn;    }    return fn;}struct Matrix2By2{    Matrix2By2    (        long long m00 = 0,         long long m01 = 0,         long long m10 = 0,         long long m11 = 0    )    :m_00(m00), m_01(m01), m_10(m10), m_11(m11)     {    }    long long m_00;    long long m_01;    long long m_10;    long long m_11;};Matrix2By2 MatrixMultiply(    const Matrix2By2& matrix1,     const Matrix2By2& matrix2){    return Matrix2By2(        matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,        matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,        matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,        matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);}Matrix2By2 MatrixPower(unsigned int n){    assert(n > 0);    Matrix2By2 matrix;    if(n == 1)    {        matrix = Matrix2By2(1, 1, 1, 0);    }    else if(n % 2 == 0)    {        matrix = MatrixPower(n / 2);        matrix = MatrixMultiply(matrix, matrix);    }    else if(n % 2 == 1)    {        matrix = MatrixPower((n - 1) / 2);        matrix = MatrixMultiply(matrix, matrix);        matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));    }    return matrix;}long long Fibonacci3(unsigned int n){    int result[2] = {0, 1};    if(n < 2)        return result[n];    Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);    return PowerNMinus2.m_00;}// ====================测试代码====================void Test(int n, int expected){    if(Fibonacci1(n) == expected)        printf("Test for %d in solution1 passed.\n", n);    else        printf("Test for %d in solution1 failed.\n", n);    if(Fibonacci2(n) == expected)        printf("Test for %d in solution2 passed.\n", n);    else        printf("Test for %d in solution2 failed.\n", n);    if(Fibonacci3(n) == expected)        printf("Test for %d in solution3 passed.\n", n);    else        printf("Test for %d in solution3 failed.\n", n);}int main(int argc, char* argv[]){    Test(0, 0);    Test(1, 1);    Test(2, 1);    Test(3, 2);    Test(4, 3);    Test(5, 5);    Test(6, 8);    Test(7, 13);    Test(8, 21);    Test(9, 34);    Test(10, 55);    Test(40, 102334155);    return 0;}

面试题10: 二进制中1的个数

很多的解法.

方法1:

不断的除以2或右移. 判断其中1的个数.

存在的问题是, 对于负数就不行了.

方法2:

将1不断左移, 然后与原数进行按位与操作.

方法3:

利用n&(n-1)的技巧

方法2和3的代码如下:

int NumberOf1_Solution1(int n){    int count = 0;    unsigned int flag = 1;    while(flag)    {        if(n & flag)            count ++;        flag = flag << 1;    }    return count;}int NumberOf1_Solution2(int n){    int count = 0;    while (n)    {        ++ count;        n = (n - 1) & n;    }    return count;}void Test(int number, unsigned int expected){    int actual = NumberOf1_Solution1(number);    if(actual == expected)        printf("Solution1: Test for %p passed.\n", number);    else        printf("Solution1: Test for %p failed.\n", number);    actual = NumberOf1_Solution2(number);    if(actual == expected)        printf("Solution2: Test for %p passed.\n", number);    else        printf("Solution2: Test for %p failed.\n", number);    printf("\n");}int main(int argc, char* argv[]){    // 输入0，期待的输出是0    Test(0, 0);    // 输入1，期待的输出是1    Test(1, 1);    // 输入10，期待的输出是2    Test(10, 2);    // 输入0x7FFFFFFF，期待的输出是31    Test(0x7FFFFFFF, 31);    // 输入0xFFFFFFFF（负数），期待的输出是32    Test(0xFFFFFFFF, 32);    // 输入0x80000000（负数），期待的输出是1    Test(0x80000000, 1);    return 0;}