hjhomw的博客

代码：#include <iostream>using namespace std;void FindSequentialNumbersWithSum(int* array, int nLength,int sum){ int first_index = 0; int last_index = 1; while (first_index <= last_index && la

代码：//二叉搜索树的后序遍历bool postOrderTraversalOfBST(int array[], int length){ if (array == nullptr || length <= 0) return false; int root = array[length - 1]; int i = 0; for (; i < le

代码：#include <iostream>#include <cstdlib>using namespace std;//打印一圈矩阵void PrintMatrixCircle(int** numbers, int columns, int rows, int start){ int endX = columns - 1 - start; int endY = rows -

代码:struct BinaryTreeNode{ int m_nValue; BinaryTreeNode* m_pLeft; BinaryTreeNode* m_pRight;};void MirrorRecursive(BinaryTreeNode* pTree){ if (pTree == nullptr || (pTree->m_pLeft == nu

代码：//结点struct ListNode{ int m_val; ListNode* m_pNext;};//第一种方法，需要两个辅助栈ListNode* FirstCommonNode(ListNode* & pList1, ListNode* & pList2){ if (pList1 == nullptr || pList2 == nullptr)

代码:struct ListNode{ int m_nValue; ListNode* m_pNext;};ListNode* ReverseList(ListNode *pHead){ ListNode* pReverseList = nullptr; ListNode* pCur = pHead; ListNode* pPrev = nullptr;

代码：struct ListNode{ int m_nValue; ListNode* m_pNext;};ListNode* findKthtoTail(ListNode* pHead, int k){ if (pHead == nullptr || k <= 0) return nullptr; ListNode *p1 = pHead, *

代码：#include <iostream>using namespace std;//逆序对int InversePairCore(int* pArray, int* pCopy, int start, int end){ if (start == end) { pCopy[start] = pArray[start]; return 0;

代码：#include <iostream>using namespace std;//判断是否为奇数bool isOdd(int num){ return (num & 1) == 1;}void frontOddbackEven(int *pData, int length, bool(*fun)(int)){ if (pData == nullptr || lengt

代码：#include <iostream>using namespace std;//打印数字void printNumber(char* number) //前面为0的数字不打印{ int len = strlen(number); int i = 0; while (number[i] == '0') i++; for (int j = i

代码：#include <iostream>#include <stack>using namespace std;template <typename T> class StackWithMin{public: StackWithMin(); ~StackWithMin(); const T& top() const; void push(const T& v

代码：#include <iostream>#include <stack>using namespace std;bool StackPushPopOrder(const int* pPush, const int* pPop, const int length){ bool bPossible = false; if (pPush == nullptr || pPop ==

代码：#include <iostream>using namespace std;void FindTwoNumbersWithSum(int* Array, int nLength, int* number1, int* number2, int sum){ int pHead = 0; int pEnd = nLength - 1; while (pHead < pE

代码：#include <iostream>using namespace std;int FindFirstIndexOf1(int number){ int index_bits = 0; while (((number & 1) == 0) && (index_bits < 8 * sizeof(int))) { number = number >>

代码：//求二叉树的深度int BSTDepth(TreeNode* pRoot){ if (pRoot == nullptr) return 0; int LeftDepth = BSTDepth(pRoot->m_pLeft); int RightDepth = BSTDepth(pRoot->m_pRight); return LeftDept

代码：#include <iostream>using namespace std;//找到第一个k的下标int GetFirstIndex(int* pArray,int k, int start, int end){ int mid = 0; while (start <= end)

代码：#include <iostream>#include <vector>using namespace std;void Permutation(char* pStr, char* pBegin){ if ( *pBegin == '\0') { cout << pStr << "\t"; } else { for (

代码：struct BinaryTreeNode{ int m_val; BinaryTreeNode* m_pLeft; BinaryTreeNode* m_pRight;};//递归方法BinaryTreeNode* BSTConvertToTwoWayList(BinaryTreeNode* pRoot){ BinaryTreeNode* pLastNo

代码：//复制复杂链表struct ComplexListNode{ int value; ComplexListNode* pNext; ComplexListNode* pSibling;};//复制链表void CopyList(ComplexListNode* pComplextListHead){ ComplexListNode* pNode =

代码：struct BinaryTreeNode{ int m_value; BinaryTreeNode* m_pLeft; BinaryTreeNode* m_pRight;};void findPath(BinaryTreeNode* pRoot, int sum, vector<int>& path, int currentSum){ currentSu

代码：struct BinaryTreeNode{ int m_val; BinaryTreeNode* m_pLeft; BinaryTreeNode* m_pRight;};void PrintUpToBottom(BinaryTreeNode* pTree){ queue<BinaryTreeNode*> treeQue; if (pTree ==

代码：#include <iostream>using namespace std;char FindFirstNotRepeatingChar(char* pString){ if (pString == nullptr) return '\0'; const int size = 256; //char 是8个bit的类型，总共有256个字符 int

代码：#include <iostream>using namespace std;//找出三个数中的最小数int min(int number1, int number2, int number3) { int min = (number1 < number2) ? number1 : number2; return (min < number3) ? min : numbe

include using namespace std;bool equal(double num1, double num2) { if ((num1 - num2> -0.0000001) && (num1-num2 < 0.0000001)) return true;return false;}double PowerExponent(double base,

代码：#include <iostream>using namespace std;int compare(const void* arg1, const void* arg2){ return *(int*)arg1 - *(int*)arg2;}bool IsContinusCards(int* pCards, int nLength){ if (pCards == nul

分析： 动态规划的思想： 1.现在变量有：骰子个数，点数和。当有k个骰子，点数和为n时，出现次数记为f(k,n)。那与k-1个骰子阶段之间的关系是怎样的？ 2.当我有k-1个骰子时，再增加一个骰子，这个骰子的点数只可能为1、2、3、4、5或6。那k个骰子得到点数和为n的情况有： (k-1,n-1)：第k个骰子投了点数1 (k-1,n-2)：第k个骰子投了点数2 (k-1,n-3)：第k个

代码：#include <iostream>using namespace std; //翻转字符数组void Reverse(char* pBegin, char* pEnd){ if (pBegin == nullptr || pEnd == nullptr) return; while (pBegin < pEnd) { char te

代码：#include <iostream>using namespace std; //翻转字符数组void Reverse(char* pBegin, char* pEnd){ if (pBegin == nullptr || pEnd == nullptr) return; while (pBegin < pEnd) { char te

解法一： 若输入数组可以修改，可以基于快速排序的思想找数组 第k个大小的数的位置，返回[0~k-1]位置的数即可。解法二：typedef multiset<int, greater<int>> intSet;typedef multiset<int, greater<int>>::iterator iterSet;void GetLeastKNumbers(const vector<int>&

代码：#include <iostream>#include <random>using namespace std;int RandomRange(int start, int end){ uniform_int_distribution<int> u(start, end); default_random_engine e; return u(e);}int Part

代码：#include <iostream>using namespace std;//八皇后void eightQueue(int* colunmnIndex, int k ,int length){ if (k == length) { for (int i = 0 ; i < length; i++) { for (

题目：用两个栈实现一个队列。队列的声明如下，请实现它的两个函数appendTail和deleteHead,分别完成在队列尾部插入结点和在队列头部删除结点的功能。代码：#include <iostream>#include <stack>#include <exception>using namespace std;template<typename T> class CQueue{publi

题目：请实现一个函数，把字符串中的每一个空格替换成”%20”。例如输入”we are happy.”，则输出”we%20are%20happy.”。分析： 将长度为1的空格替换为长度为3的“%20”，字符串的长度变长。如果允许我们开辟一个新的数组来存放替换空格后的字符串，那么这道题目就非常简 单。设置两个指针分别指向新旧字符串首元素，遍历原字符串，如果碰到空格就在新字符串上填入“%20”，否则就复

代码：#include <iostream>#include <list>using namespace std;int FindLastNumber(int n, int m){ if (n < 1 || m < 1) return -1; //用list模拟循环链表 list<int> numbers; for (int i = 0; i < n;

代码：#include <iostream>#include <list>using namespace std;//方法一，利用构造函数求解class A{public: A() { ++n; sum += n; } static void Reset() { n = 0; sum

代码：#define _CRT_SECURE_NO_WARNINGS#include <iostream>using namespace std;const int g_max_length = 10; // 0~9char* g_str_combine1 = new char[2 * g_max_length + 1];char* g_str_combine2 = new char[2 *

代码：#include <iostream>using namespace std;/*//右移有可能导致死循环int Numberof1(int n) //循环次数等于整数二进制的位数{int count = 0;while (n){if (n&1)count++;n = n>>1;}return count;}//常规解法int Numberof1(int n) //

代码：#include <iostream>using namespace std;int MinInOrder(int num[], int index1, int index2){ int min_num = num[index1]; while (index1 < index2) { ++index1; if (min_num > nu

代码：#include <iostream>using namespace std;//迭代求解斐波那契数列int Fibonacci(int n){ if (n <= 0) throw new exception("Invalid Parameters"); if (n == 1 || n == 2) return 1; int Fib_1

代码：#define _CRT_SECURE_NO_WARNINGS#include <iostream>using namespace std;int PowerBase10(int n){ int result = 1; for (int i = 0; i < n; i++) { result *= 10; } return resu