Chap2: question: 1

1. 赋值运算符函数（或应说复制拷贝函数问题）

class A

{

private:

	int value;

public:

	A(int n) : value(n) {}

	A(A O) { value = O.value; } // Compile Error : (const A& O)

};

因为，A a(0); A b = a; 就会使程序陷入死循环。

2. 实现 Singleton 模式 (C#)

（博客待加设计模式总结）

3.二维数组中的查找

Sample:

二维数组：Matrix[4][4]，行列都是递增。

1 　　2　　 8　　 9

2 　　4　　 9　　12

4 　　7 　　10 　 13

6 　　8 　　11 　 15

判断 value = 7 是否在数组。

思路：从右上角开始，若大于 7 删去一列；若小于 7 删去一行。

代码：

#include<iostream>

const int N = ;

int data[][N] = {{, , , },{ , , , },{, , , }, {, , , }};

bool find(int (*matrix)[N], int row, int column, int value)

{

    int r  = , c = column - ;

    while(r < row && c >= )

    {

        if(matrix[r][c] == value) return true;

        if(matrix[r][c] > value) --c;

        else ++r;

    }

    return false;

}

int main()

{

    std::cout << find(data, , , ) << std::endl;

    return ;

}

4.替换空格时间：O(n) 空间：O(1)

Sample:

输入: S:"We are happy."

输出：S:"We%20are%20happy."

 #include<stdio.h>

 #include<string.h>

 const int N = ;

 /* length  is the full capacity of the string, max number of elements is length-1*/

 void ReplaceBank(char s[], int length){

     if(s == NULL && length <= ) return; // program robustness

     int nowLength = -, numberOfBlank = ;

     while(s[++nowLength] != '\0'){

         if(s[nowLength] == ' ') ++numberOfBlank;

     }

     int newLength = nowLength + numberOfBlank * ;  // newLength is the number of elements

     if(newLength >= length) return;

     int idOfNow = nowLength, idOfNew = newLength; // both point to '/0'

     while(idOfNow >= ){      /* key program */

         if(s[idOfNow] == ' ') {

             strncpy(s+idOfNew-, "%20", );

             idOfNew -= ;

             --idOfNow;

         }else

             s[idOfNew--] = s[idOfNow--];

     }

 }

 int main(){

     char s[N] = "We are happy.";

     puts(s);

     ReplaceBank(s,N);

     puts(s);

     return ;

 }

Code

5.从尾到头打印链表

1. 询问是否可以改变链表结构，若可以，则空间O(1)；否则，

2. 使用栈，或者递归。

a. 使用栈：

 #include <iostream>

 #include <string>

 #include <stack>

 struct LinkNode{

     char e;

     LinkNode *next;

 };

 void print(LinkNode *Head)

 {

     std::stack<char> st;

     while(Head != NULL)

     {

         st.push(Head->e);

         Head = Head->next;

     }

     while(!st.empty())

     {

         printf("%c ", st.top());

         st.pop();

     }

     printf("\n");

 }

 LinkNode* init(const std::string &s)

 {

     size_t i = ;

     LinkNode *head, *p;

     p = head = NULL;

     while(i < s.length())

     {

         LinkNode *p2 = new LinkNode;

         p2->e = s[i++]; p2->next = NULL;

         if(head == NULL)

             head = p = p2;

         else

         {

             p->next = p2;

             p = p->next;

         }

     }

     return head;

 }

 void release(LinkNode *Head){

     if(Head == NULL) return;

     release(Head->next);

     delete[] Head;

     Head = NULL;

 }

 int main()

 {

     const std::string s = "ABCDEFG";

     LinkNode *Head = init(s);

     print(Head);

     release(Head);

     return ;

 }

Code

b. 使用递归:

void RecursivePrint(LinkNode *Head)

{

    if(Head == NULL) return;

    RecursivePrint(Head->next);

    printf("%c ", Head->e);

}

6. 重建二叉树

　　a. 前序 && 中序

 #include <cstdio>

 #include <queue>

 struct BTNode{

     int value;

     BTNode *pLeft;

     BTNode *pRight;

     BTNode(int x) : value(x), pLeft(NULL), pRight(NULL) {}

 };

 BTNode* constructCore(int *startPreOrder, int *endPreOrder, int *startInOrder, int *endInOrder)

 {

     int rootValue = startPreOrder[];

     BTNode *root = new BTNode(rootValue);

     int *rootInOrder = startInOrder;

     while(*rootInOrder != rootValue && rootInOrder <= endInOrder) ++rootInOrder;

     if(*rootInOrder != rootValue) { printf("Invalid Input!"); return NULL; }

     int leftLength = rootInOrder - startInOrder;

     if(leftLength > )

     {

         root->pLeft = constructCore(startPreOrder + , startPreOrder + leftLength,

             startInOrder, startInOrder + leftLength - );

     }

     if(rootInOrder != endInOrder)

     {

         root->pRight = constructCore(startPreOrder + leftLength + , endPreOrder,

             rootInOrder + , endInOrder);

     }

     return root;

 }

 BTNode* construct(int *preOrder, int *inOrder, int length)

 {

     if(preOrder == NULL || inOrder == NULL || length <= )

         return NULL;

     return constructCore(preOrder, preOrder + length - , inOrder, inOrder + length -);

 }

 void print(BTNode *root)

 {

     if(root == NULL) return;

     std::queue<BTNode*> qu;

     qu.push(root);

     while(!qu.empty())

     {

         BTNode *p = qu.front();

         qu.pop();

         if(p->pLeft) qu.push(p->pLeft);

         if(p->pRight) qu.push(p->pRight);

         printf("%d ", p->value);

     }

 }

 void release(BTNode *root)

 {

     if(root == NULL) return;

     release(root->pLeft);

     release(root->pRight);

     delete[] root;

     root = NULL;

 }

 int main()

 {

     int preOrder[] = {, , , , , , , };

     int inOrder[] = {, , , , , , , };

     BTNode *root = construct(preOrder, inOrder, );

     print(root);

     release(root);

     return ;

 }

Code

　　　　二叉树的各种遍历：

 #include <cstdio>

 #include <queue>

 #include <stack>

 struct BTNode{

     int value;

     BTNode *pLeft;

     BTNode *pRight;

     BTNode(int x) : value(x), pLeft(NULL), pRight(NULL) {}

 };

 BTNode* constructCore(int *startPreOrder, int *endPreOrder, int *startInOrder, int *endInOrder)

 {

     int rootValue = startPreOrder[];

     BTNode *root = new BTNode(rootValue);

     int *rootInOrder = startInOrder;

     while(*rootInOrder != rootValue && rootInOrder <= endInOrder) ++rootInOrder;

     if(*rootInOrder != rootValue) { printf("Invalid Input!\n"); return NULL; }

     int leftLength = rootInOrder - startInOrder;

     if(leftLength > )

     {

         root->pLeft = constructCore(startPreOrder + , startPreOrder + leftLength,

             startInOrder, startInOrder + leftLength - );

     }

     if(rootInOrder != endInOrder)

     {

         root->pRight = constructCore(startPreOrder + leftLength + , endPreOrder,

             rootInOrder + , endInOrder);

     }

     return root;

 }

 BTNode* construct(int *preOrder, int *inOrder, int length)

 {

     if(preOrder == NULL || inOrder == NULL || length <= )

         return NULL;

     return constructCore(preOrder, preOrder + length - , inOrder, inOrder + length -);

 }

 void levelOrderPrint(BTNode *root)

 {

     if(root == NULL) return;

     printf("BFS：");

     std::queue<BTNode*> qu;

     qu.push(root);

     while(!qu.empty())

     {

         BTNode *p = qu.front();

         qu.pop();

         if(p->pLeft) qu.push(p->pLeft);

         if(p->pRight) qu.push(p->pRight);

         printf("%-3d ", p->value);

     }

     printf("\n");

 }

 void preORderPrint2(BTNode *root) // preORder

 {

     if(root == NULL) return;

     printf("DLR: ");

     std::stack<BTNode*> st;

     st.push(root);

     while(!st.empty())

     {

         BTNode *p = st.top();

         st.pop();

         if(p->pRight) st.push(p->pRight);

         if(p->pLeft) st.push(p->pLeft);

         printf("%-3d ", p->value);

     }

     printf("\n");

 }

 void inOrderPrint3(BTNode *root) // inOrder

 {

     if(root == NULL) return;

     printf("LDR: ");

     std::stack<BTNode*> st;

     st.push(root);

     BTNode *p = root;

     while(p->pLeft)

     {

         st.push(p->pLeft);

         p = p->pLeft;

     }

     while(!st.empty())

     {

         p = st.top();

         st.pop();

         if(p->pRight)

         {

             BTNode *q = p->pRight;

             st.push(q);

             while(q->pLeft) { st.push(q->pLeft); q = q->pLeft; }

         }

         printf("%-3d ", p->value);

     }

     printf("\n");

 }

 void postOrderPrint4(BTNode *root) // postOrder

 {

     if(root == NULL) return;

     printf("LRD: ");

     std::stack<BTNode*>st;

     st.push(root);

     BTNode *p = root;

     while(p->pLeft || p->pRight)

     {

         while(p->pLeft) { st.push(p->pLeft); p = p->pLeft; }

         if(p->pRight) { st.push(p->pRight); p = p->pRight; }

     }

     //bool tag = true;

     while(!st.empty())

     {

         BTNode *q = st.top();

         st.pop();

         if(!st.empty())

         {

             p = st.top();

             if(p->pRight && p->pRight != q)

             {

                 st.push(p->pRight); p = p->pRight;

                 while(p->pLeft || p->pRight)

                 {

                     while(p->pLeft) { st.push(p->pLeft); p = p->pLeft; }

                     if(p->pRight) { st.push(p->pRight); p = p->pRight; }

                 }

             }

         }

         printf("%-3d ", q->value);

     }

     printf("\n");

 }

 void DFSPrint(BTNode *root,const int nVertex)

 {

     if(root == NULL) return;

     printf("DFS: ");

     bool *visit = new bool[nVertex];

     memset(visit, false, nVertex);

     std::stack<BTNode*> st;

     st.push(root);

     visit[root->value] = true;

     while(!st.empty())

     {

         BTNode *p = st.top();

         st.pop();

         if(p->pRight && !visit[p->pRight->value]) { st.push(p->pRight); visit[p->pRight->value] = true; }

         if(p->pLeft && !visit[p->pLeft->value]) { st.push(p->pLeft); visit[p->pLeft->value];}

         printf("%-3d ", p->value);

     }

     printf("\n");

 }

 void release(BTNode *root)

 {

     if(root == NULL) return;

     release(root->pLeft);

     release(root->pRight);

     delete[] root;

     root = NULL;

 }

 int main()

 {

     int preOrder[] = {, , , , , , , , , , , , , , };

     int inOrder[] = {, , , , , , , , , , , , , , };

     BTNode *root = construct(preOrder, inOrder, );

     levelOrderPrint(root);

     preORderPrint2(root);

     inOrderPrint3(root);

     postOrderPrint4(root);

     DFSPrint(root, +);

     release(root);

     return ;

 }

Code

7.用两个栈实现队列

 #include <cstdio>

 #include <stack>

 #include <exception>

 template<typename T> class CQueue

 {

 public:

     void appendTail(T x);

     T deleteHead();

 private:

     std::stack<T> st1;

     std::stack<T> st2;

 };

 template<typename T>void CQueue<T>::appendTail(T x)

 {

     st1.push(x);

 }

 template<typename T>T CQueue<T>::deleteHead()

 {

     if(st2.empty())

     {

         if(st1.empty()) throw new std::exception("queue is empty!");

         while(!st1.empty())

         {

             st2.push(st1.top());

             st1.pop();

         }

     }

     T v = st2.top();

     st2.pop();

     return v;

 }

 int main()

 {

     printf("Test int:\n");

     CQueue<int> qu;

     qu.appendTail();

     qu.appendTail();

     printf("%d\n",qu.deleteHead());

     printf("%d\n",qu.deleteHead());

     printf("Test char*:\n");

     CQueue<char*> qu2;

     qu2.appendTail("Hello");

     qu2.appendTail("World!");

     printf("%s\n",qu2.deleteHead());

     printf("%s\n",qu2.deleteHead());

     //printf("%s\n",qu2.deleteHead());

     return ;

 }

Code

8.旋转数组的最小数字

 #include <iostream>

 using namespace std;

 int MinInorder(int *numbers, int low, int high)

 {

     int minNum = numbers[low];

     while(++low <= high)

     {

         if(numbers[low] < minNum)

             minNum = numbers[low];

     }

     return minNum;

 }

 int Min(int *numbers, int length)

 {

     if(numbers == NULL || length <= ) throw new exception("Invalid parameters!");

     int low = , high = length - ;

     int mid = low;   // note

     while(numbers[low] >= numbers[high]) // note >=

     {

         if(high - low == )

         {

             mid = high;

             break;

         }

         mid = (low + high) >> ;

         if(numbers[mid] == numbers[low] && numbers[mid] == numbers[high]) return MinInorder(numbers, low, high);

         else if(numbers[mid] >= numbers[low]) low = mid;

         else if(numbers[mid] <= numbers[high]) high = mid;

     }

     return numbers[mid];

 }

 int main()

 {

     int test1[] = {, , , , };

     cout << "Test1: " << Min(test1, sizeof(test1)/) << endl;

     int test2[] = {, , , , };

     cout << "Test2: " << Min(test2, sizeof(test2)/) << endl;

     return ;

 }

Code

9.斐波那契数列第 n 项

a. 动态规划（从低到高保存结果）

int Fibonacci1(unsigned long long N)

{

	long long fibN;

	long long a[] = {0, 1};

	if(N < 2) return a[N];

	while(N >= 2)

	{

		fibN = (a[0] + a[1]) % M;

		a[0] = a[1];

		a[1] = fibN;

		--N;

	}

	return (int)fibN;

}

b1.矩阵二分乘（递归）

const int M = 1e+9 +7 ;

struct Matrix{

	long long e[2][2];

};

Matrix Mat;

Matrix multiplyMatrix(Matrix &A, Matrix &B)

{

	Matrix C;

	for(int i = 0; i < 2; ++i){

		for(int j = 0; j < 2; ++j){

			C.e[i][j] = ((A.e[i][0] * B.e[0][j]) % M + (A.e[i][1] * B.e[1][j]) % M) % M;

		}

	}

	return C;

}

Matrix getMatrix(long long n)

{

	if(n == 1) return Mat;

	Matrix tem = getMatrix(n>>1);

	tem = multiplyMatrix(tem,tem);

	if((n & 0x1) == 0) return tem;

	else

		return multiplyMatrix(Mat,tem);

}

int Fibonacci2(long long N)

{

	if(N == 0) return 0;

	if(N == 1) return 1;

	Mat.e[0][0] = 1; Mat.e[0][1] = 1;

	Mat.e[1][0] = 1; Mat.e[1][1] = 0;

	Matrix result = getMatrix(N-1);

	return (int)result.e[0][0];

}

b2. 矩阵二分乘(非递归)

const int M = 1000000007;

struct Matrix{

	long long e[2][2];

};

Matrix Mat;

Matrix multiplyMatrix(Matrix &A, Matrix &B)

{

	Matrix C;

	for(int i = 0; i < 2; ++i){

		for(int j = 0; j < 2; ++j){

			C.e[i][j] = ((A.e[i][0] * B.e[0][j]) % M + (A.e[i][1] * B.e[1][j]) % M) % M;

		}

	}

	return C;

}

Matrix getMatrix(Matrix base, long long N)

{

	Matrix T;                   // set one unit matrix

	T.e[0][0] = T.e[1][1] = 1;

	T.e[0][1] = T.e[1][0] = 0;

	while(N - 1 != 0)

	{

		if(N & 0x1) T = multiplyMatrix(T, base);

		base = multiplyMatrix(base, base);

		N >>= 1;

	}

	return multiplyMatrix(T, base);

}

int Fibonacci2(long long N)

{

	if(N == 0) return 0;

	if(N == 1) return 1;

	Mat.e[0][0] = 1; Mat.e[0][1] = 1;

	Mat.e[1][0] = 1; Mat.e[1][1] = 0;

	Matrix result = getMatrix(Mat, N-1);

	return (int)result.e[0][0];

}

两种方法效率的比较：

 #include <iostream>

 #include <ctime>

 using namespace std;

 const int M = ;

 struct Matrix{

     long long e[][];

 };

 Matrix Mat;

 Matrix multiplyMatrix(Matrix &A, Matrix &B)

 {

     Matrix C;

     for(int i = ; i < ; ++i){

         for(int j = ; j < ; ++j){

             C.e[i][j] = ((A.e[i][] * B.e[][j]) % M + (A.e[i][] * B.e[][j]) % M) % M;

         }

     }

     return C;

 }

 Matrix getMatrix(Matrix base, long long N)

 {

     Matrix T;                   // set one unit matrix

     T.e[][] = T.e[][] = ;

     T.e[][] = T.e[][] = ;

     while(N -  != )

     {

         if(N & 0x1) T = multiplyMatrix(T, base);

         base = multiplyMatrix(base, base);

         N >>= ;

     }

     return multiplyMatrix(T, base);

 }

 int Fibonacci2(long long N)

 {

     if(N == ) return ;

     if(N == ) return ;

     Mat.e[][] = ; Mat.e[][] = ;

     Mat.e[][] = ; Mat.e[][] = ;

     Matrix result = getMatrix(Mat, N-);

     return (int)result.e[][];

 }

 int Fibonacci1(long long N)

 {

     long long fibN;

     long long a[] = {, };

     if(N < ) return (int)a[N];

     while(N >= )

     {

         fibN = (a[] + a[]) % M;

         a[] = a[];

         a[] = fibN;

         --N;

     }

     return (int)fibN;

 }

 int main()

 {

     unsigned long long N;

     while(cin >> N)

     {

         /* method 1: DP  */

         clock_t start = clock();

         int fibN = Fibonacci1(N);

         clock_t end = clock();

         cout << "method1：" << fibN << " time: " << end-start << "ms" << endl;

         /* method 2   */

         start = clock();

         fibN = Fibonacci2(N);

         end = clock();

         cout << "method2：" << fibN << " time: " << end-start << "ms" << endl;

     }

     return ;

 }

Code

10.一个整数的二进制表示中 1 的个数。（包含正负数）

需要注意的是：—1 >> (任意位) == —1；负数 >> +∞ == —1。（（—1）₁₀ == （0xffffffff）₁₆ ）

a. 利用辅助变量，每次判断 n 的一位。

 int numberOf1(int n)

 {

     int count = ;

     int flag = ;

     while(flag)

     {

         if(n & flag) count ++;

         flag <<= ;

     }

     return count;

 }

Code

b. 利用 n 的性质。

 #include <iostream>

 int numberOf1(int n){

     int count = ;

     while(n){

         count ++;

         n &= (n-);

     }

     return count;

 }

 int main(){

     int N;

     while(std::cin >> N){

         std::cout << "has 1: " << numberOf1(N) << std::endl;

     }

     return ;

 }

Code

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