C++常见笔试编程题

素颜马尾好姑娘i 2022-11-12 01:43 221阅读 0赞

1从尾到头打印单链表

//栈的方式
    void printListReversingly_iteratively(LinkNode* pHead)
    {
        
    	std::stack<LinkNode*> nodes;
    	LinkNode* pNode = pHead;
    	while (pHead!=nullptr) 
    	{
        
    		nodes.push(pNode);
    		pNode = pNode->m_pNext;
    	}
    	while (!nodes.empty())
    	{
        
    		pNode = nodes.top();
    		std::cout << pNode->m_value << " ";
    		nodes.pop();
    	}
    }
    //递归
    void printListReversingly_recursively(LinkNode* pHead)
    {
        
    	if (pHead != nullptr)
    	{
        
    		if (pHead->m_pNext != nullptr)
    		{
        
    			printListReversingly_recursively(pHead->m_pNext);
    		}
    		std::cout << pHead->m_value << " ";
    	}
    }

2反转单链表

Node* ReverseList(Node* pHead)
    {
        
    	Node* pReverse = NULL;
    	Node* pNode = pHead;
    	Node* pPrev = NULL;
    	while (pNode!=NULL)
    	{
        
    		Node* pNext = pNode->next;
    		if (pNext == NULL)
    			pReverse = pNode;
    		pNode->next = pPrev; 
    		pPrev = pNode;
    		pNode = pNext;
    	}
    	return pReverse;
    }

Node* ReverseList2(Node* pHead)
    {
        
    	if (pHead ==NULL || pHead->next == NULL )
    		return pHead;
    	Node* pre = pHead;
    	Node* cur = pHead->next;
    	pre->next = NULL;
    	while (cur != NULL)
    	{
        
    		Node* next = cur->next;
    		cur->next = pre;
    		pre = cur;
    		cur = next;
    	}
    	return pre;
    }

Node* ReverseList3(Node* pHead)
    {
        
    	if (pHead == NULL || pHead->next == NULL)
    		return pHead;//递归条件终止，找到链表的最后一个结点
    	Node* newHead = ReverseList3(pHead->next);//先递后归，从后往前变量每个节点进行反转
    	pHead->next->next = pHead;//将当前结点的后一个结点的next指向当前结点
    	pHead->next = NULL;//断开当前结点指向后一个结点
    	return newHead;
    }

Node* ReverseList4(Node* pHead)
    {
        
    	if (pHead == NULL || pHead->next == NULL)
    		return pHead;
    	Node* newHead = NULL;
    	while (pHead != NULL)
    	{
        
    		//对之前的链表头删
    		Node* node = pHead;
    		pHead = pHead->next;
    		//对新链表做头插
    		node->next = newHead;
    		newHead = node;
    	}
    	return newHead;
    }

Node* ReverseList5(Node* pHead)
    {
        
    	if (pHead == NULL || pHead->next == NULL)
    		return pHead;
    	Node* p0 = NULL;
    	Node* p1 = pHead;
    	Node* p2 = pHead->next;
    	while (p1 != NULL)
    	{
        
    		p1->next = p0;
    		p0 = p1;
    		p1 = p2;
    		if (p2 != NULL)
    			p2 = p2->next;
    	}
    	return p0;
    }

3完数：1000以内的完数，6的因子1,2,3 1+2+3=6

#include<stdio.h>
    #include<stdlib.h>
    int main()
    {
        
    	int i, j, s;//i为可能的完数，j可能的因子，s因子和
    	printf("1000以内的完数：\n");
    	for (i = 2; i <= 1000; i++)
    	{
        
    		s = 0;//保证每次循环是s的初始值为0
    		for (j = 1; j < i; j++)//为其本身
    		{
        
    			if (i%j == 0)//判断j是否为i的因子
    				s += j;
    		}
    		if (s == i)//判断因子和原数是否相等
    			printf("%3d是完数\n", i);
    	}
    	system("pause");
    	return 0;
    }