1. 寻找两个有序数组的中位数
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
double ret = -1;
vector<double> buff;
//合并两个数组
for (auto e : nums1)
buff.push_back(e);
for (auto a : nums2)
buff.push_back(a);
//将合并后的结果进行排序
sort(buff.begin(), buff.end());
int size3 = buff.size();
//获取中位数
if (size3 % 2 == 0)
{
ret = ((buff[size3 / 2] + buff[size3 / 2 - 1]) / 2);
}
else
{
ret = buff[size3 / 2];
}
return ret;
}
2. C++实现线程安全的单例模式
class singleton
{
protected:
singleton()
{
pthread_mutex_init(&mutex);
}
private:
static singleton* p;
public:
static pthread_mutex_t mutex;
static singleton* initance();
};
pthread_mutex_t singleton::mutex;
singleton* singleton::p = NULL;
singleton* singleton::initance()
{
if (p == NULL)
{
pthread_mutex_lock(&mutex);
if (p == NULL)
p = new singleton();
pthread_mutex_unlock(&mutex);
}
return p;
}
3. 数组中有三个数字出现超过1/4，求这三个数字？
#include <iostream>
using namespace std;
//求x!中k因数的个数。
int Grial(int x,int k)
{
int Ret = 0;
while (x)
{
Ret += x / k;
x /= k;
}
return Ret;
}
int main()
{
cout << Grial(10, 2) << endl;
return 0;
}
//假设要求一个n!中k的因子个数，那么必然满足例如以下的规则。
//即x=n/k+n/k^2+n/k^3...(直到n/k^x小于0);
#include <iostream>
using namespace std;
int Grial(int x, int k)
{
int count = 0;
int n = x;
while (n)
{
n &= (n - 1);
count++;
}
return x - count;
}
int main()
{
cout << Grial(3, 2) << endl;
return 0;
}
//找出数组中出现次数超过数组一半的数字。
#include <iostream>
using namespace std;
int Grial(int a[], int n)
{
int count=0;
int key;
for (int i = 0; i < n; i++)
{
if (count == 0)
{
key = a[i];
count = 1;
}
else
{
if (key == a[i])
{
count++;
}
else
{
count--;
}
}
}
return key;
}
int main()
{
int a[] = {1,2,3,4,5,6,3,3,3,3,3};
cout<<Grial(a, sizeof(a) / sizeof(int))<<endl;
return 0;
}
#include <iostream>
//上一题的扩展，有3个数字出现次数超过1/4。
using namespace std;
void Grial(int a[], int n)
{
if (n <= 3)return;
int count1=0, key1=0;
int count2=0, key2=0;
int count3=0, key3=0;
for (int i = 0; i < n; i++)
{
if (!count1 && key2 != a[i] && key3 != a[i])
{
count1++;
key1 = a[i];
}
else if (key1 == a[i])
{
count1++;
}
else if (key2!=a[i] && key3!=a[i])
{
count1--;
}
if (!count2 &&key3 != a[i] && key1!=a[i])
{
count2++;
key2 = a[i];
}
else if (key2 == a[i])
{
count2++;
}
else if (key1!=a[i] && key3!=a[i])
{
count2--;
}
if (!count3 && key1!=a[i] && key2!=a[i])
{
count3++;
key3 = a[i];
}
else if (key3 == a[i])
{
count3++;
}
else if (key1!=a[i] && key2!=a[i])
{
count3--;
}
}
cout << key1 << endl;
cout << key2 << endl;
cout << key3 << endl;
}
int main()
{
int a[] = {1,5,5,5,5,2,3,1,2,2,1,1,1,2};
Grial(a, sizeof(a) / sizeof(int));
return 0;
}
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