【Twinkle】携程2019秋招技术类笔试-后台开发工程师

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20道题目+3编程题

---

1. 统计数字二进制中1的个数

#include <iostream>
using namespace std;
int main() 
{
	long a;
		while (cin >> a)//注意while处理多个case
		{
			int c = 0;
			for (; a; ++c)
			{
				a &= (a - 1); // 清除最低位的1
			}
			cout << c<< endl;
		}
		return 0;
}

---

2.查表算法logn（我这不是这个复杂度）

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
	long count = 0; // 数据数量
	long time = 0;
	cin >> count;
	cin >> time;

	vector<vector<long>> data;
	//输入数据
	for (int i = 0; i < count; i++)
	{
		vector<long> temp;
		long t = 0;
		cin >> t;
		temp.push_back(t);
		cin >> t;
		temp.push_back(t);
		cin >> t;
		temp.push_back(t);

		data.push_back(temp);
	}

	vector<long> result;
	//判断
	for (vector<vector<long>>::iterator it = data.begin(); it != data.end(); it++)
	{
		if ((*it)[1] > time) continue;
		if ((*it)[2] < time) continue;

		result.push_back((*it)[0]);
	}

	//输出
	if (result.size() == 0)
	{
		cout << "null";
	}
	else
	{
		sort(result.begin(), result.end());
		for (int i = 0; i < result.size(); i++)
		{
			cout << result[i] << endl;
		}
	}
	return 0;
}

---

3. 这个就有点皮了

设计并实现一个LRU Cache。

#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;

// 双向链表的节点结构
struct LRUNode 
{
	int key;
	int value;
	LRUNode* prev;
	LRUNode* next;
	LRUNode() :key(0), value(0), prev(NULL), next(NULL) {}
};

class LRU
{
private:
	unordered_map<int, LRUNode*> m;  // 代替hash_map
	LRUNode* head;     // 指向双链表的头结点
	LRUNode* tail;     // 指向双链表的尾结点
	int capacity;           // Cache的容量
	int count;              // 计数
public:
	LRU(int capacity);       // 构造函数
	~LRU();                  // 析构函数
	int get(int key);             // 查询数据项
	void put(int key, int value); // 未满时插入,已满时替换
private:
	void removeLRUNode();                 // 删除尾结点（最久未使用）
	void detachNode(LRUNode* node);    // 分离当前结点
	void insertToFront(LRUNode* node); // 节点插入到头部
};

LRU::LRU(int capacity)
{
	this->capacity = capacity;
	this->count = 0;
	head = new LRUNode;
	tail = new LRUNode;
	head->prev = NULL;
	head->next = tail;
	tail->prev = head;
	tail->next = NULL;
}

LRU::~LRU()
{
	delete head;
	delete tail;
}

int LRU::get(int key)
{
	if (m.find(key) == m.end())  // 没找到
		return -1;
	else
	{
		LRUNode* node = m[key];
		detachNode(node);      // 命中，移至头部 
		insertToFront(node);
		return node->value;
	}
}

void LRU::put(int key, int value)
{
	if (m.find(key) == m.end())  // 没找到
	{
		LRUNode* node = new LRUNode;
		if (count == capacity)   // Cache已满
			removeLRUNode();

		node->key = key;
		node->value = value;
		m[key] = node;          // 插入哈希表
		insertToFront(node);    // 插入链表头部
		++count;
	}
	else
	{
		LRUNode* node = m[key];
		detachNode(node);
		node->value = value;
		insertToFront(node);
	}
}

void LRU::removeLRUNode()
{
	LRUNode* node = tail->prev;
	detachNode(node);
	m.erase(node->key);
	--count;
}

void LRU::detachNode(LRUNode* node)
{
	node->prev->next = node->next;
	node->next->prev = node->prev;
}

void LRU::insertToFront(LRUNode* node)
{
	node->next = head->next;
	node->prev = head;
	head->next = node;
	node->next->prev = node;
}

int main()
{
	int count = 0;
	cin >> count;
	LRU data(count);
	vector<int> result;			  //输出的结果

	char index = ' ';
	while (cin >> index)
	{
		if (index == 'p')
		{
			int a = 0, b = 0;
			cin >> a >> b;
			//result.push_back(data.put(a, b));
			data.put(a, b);
		}

		if (index == 'g')
		{
			int a = 0;
			cin >> a;
			result.push_back(data.get(a));
		}
	}

	for (int i = 0; i < result.size(); i++)
	{
		cout << result[i] << endl;
	}
	
	return 0;
}

参考： 设计并实现一个LRU Cache