腾讯实习生招聘技术类研发笔试题

#include <stdio.h>

int main()
{
    printf("\n");
    int a[5] = {1,2,3,4,5};
    int *p, **k;
    p = a;
    k = &p;
    printf("%d", *(p++));
    printf("%d", **k);
    return 0;
}
输出结果：
12
解释：1、p指针指向a数组的首地址，printf("%d", *(p++));时输出a[0]的值。
2、由于p++，p的值改变指向a[1]，k指向p所在地址，故*k为p中值，**k等同于*p，即a[1]。

---

#include <iostream>
using namespace std;
class Base
{
    int x;
public:
    Base(int b):x(b){}
    virtual void display()
    {
        cout << x << endl;
    }
};
class Derived : public Base
{
    int y;
public:
    Derived(int d):Base(d), y(d) {}
    void display()
    {
        cout << y << endl;
    }
};
int main(void)
{
    Base b(2);
    Derived d(3);
    b.display();
    d.display();
    Base *p = &d;
    p->display();
    system("pause");
    return 0;
}
运行结果：
2
3
3
解释：p->display();Derived继承了Base，但是p->display()调用的依然是子类Derived的display.

 

---

 

#include <stdio.h>

struct xx
{
    long long _x1;
    char _x2;
    int _x3;
    char _x4[2];
    static int _x5;
};
int xx::_x5;
int main(void)
{
    long long _x1;
    printf("x1:%d\n[0x%p]", sizeof(_x1), &xx._x1);
    char _x2;
    printf("x2:%d\n", sizeof(_x2));
    int _x3;
    printf("x3:%d\n", sizeof(_x3));
    char _x4[2];
    printf("x4:%d\n", sizeof(_x4));

    printf("x5:%d\n", sizeof(xx::_x5));
    printf("%d", sizeof(xx));
    return 0;
}
运行结果：
x1:8
x2:1
x3:4
x4:2
x5:4
24
解释：字节对齐问题，8+4（1）+4+4（2）+4=24

 

---

春节期间小明使用微信收到很多个红包，非常开心。在查看领取红包记录时发现，某个红包金额出现的次数超过了红包总数的一半。请帮小明找到该红包金额。写出具体算法思路和代码实现，要求算法尽可能高效。
#include <stdio.h>
#include <stdlib.h>
#define LEN sizeof(struct node)

double findvalue = 0.0;

struct node
{
    double value;
    int num;
    struct node * pleft;
    struct node * pright;
};
struct node * head = (struct node *)malloc(LEN);

void insertTree(double temp, struct node * p)
{
    if(p->value == 0)
    {
        p->value = temp;
    }
    else if(temp>p->value)
    {
        if(p->pright == NULL)
        {
            p->pright = (struct node *)malloc(LEN);
            p->pright->value = 0;
            p->pright->num = 1;
            p->pright->pleft = NULL;
            p->pright->pright = NULL;
        }
        insertTree(temp, p->pright);
    }
    else if(temp<p->value)
    {
        if(p->pleft == NULL)
        {
            p->pleft = (struct node *)malloc(LEN);
            p->pleft->value = 0;
            p->pleft->num = 1;
            p->pleft->pleft = NULL;
            p->pleft->pright = NULL;
        }
        insertTree(temp, p->pleft);
    }
    else if(temp == p->value)
    {
        p->num ++;
    }
}

void findTree(struct node * p, int n)
{
    if(p->pleft!=NULL) findTree(p->pleft, n);
    if(p!=NULL) 
    {
        if(p->num > n/2) findvalue = p->value;
    }
    if(p->pright!=NULL) findTree(p->pright, n);
}

int main(void)
{
    int n;//红包总数
    
    double temp;//单个红包金额
    
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%lf", &temp);
        head->value = temp;
        head->num = 1;
        head->pleft = NULL;
        head->pright = NULL;
        for(int i=1; i<n; i++)//建立相同金额的红包树形结构。value记录红包金额，num记录红包个数
        {
            scanf("%lf", &temp);
            insertTree(temp, head);
        }
        findTree(head,n);//查找相同金额数大于总数一半的红包金额
        printf("%lf", findvalue);//输出相同金额数大于总数一半的红包金额
    }
    
    return 0;
}

 

---

转换方法之递归生成码表
这种方法基于格雷码是反射码的事实，利用递归的如下规则来构造：
1、1位格雷码有两个码字
2、(n+1)位格雷码中的前2n个码字等于n位格雷码的码字，按顺序书写，加前缀0
3、(n+1)位格雷码中的后2n个码字等于n位格雷码的码字，按逆序书写，加前缀1
4、n+1位格雷码的集合 = n位格雷码集合(顺序)加前缀0 + n位格雷码集合(逆序)加前缀1

class GrayCode {
public:
    vector<string> getGray(int n) {
        // write code here
        vector<string> result;
        if(n==1)
        {
            result.push_back("0");
            result.push_back("1");
            return result;
        }
        vector<string> v = getGray(n-1);
        int len = pow(2,n-1);
        for(int i=0; i<len; i++)
        {
            result.push_back("0"+v[i]);//n位格雷码集合(顺序)加前缀0
        }
        for(int i=len-1; i>=0; i--)
        {
            result.push_back("1"+v[i]);//n位格雷码集合(逆序)加前缀1
        }
        return result;
    }
};