我正在尝试使用随机生成的数字填充向量,但问题是当我运行它时,我在排序后只得到2个数字。
0 0 0 0 0 17 17 17
这是我的代码,所以你能告诉我,我做错了吗
#include <iostream>
#include <random>
#include <functional>
using namespace std;
using namespace std::chrono;
class Rand_int {
public:
Rand_int(int low, int high) :dist{low,high} {
for (int i = 0; i < 100; i++)
cout << "Generated: " << dist(re) << endl;
num = dist(re); }
int operator()() { return dist(re);}
int getNum() const
{
return num;
}
bool operator()(int x) const
{
return (num <= x);
}
private:
default_random_engine re;
uniform_int_distribution<> dist;
int num;
};
int main()
{
int n=10;
vector<int> v(n);
Rand_int rnd {0,20}; // make a uniform random number generator
for (int i=0; i<n; i++)
{
vector<int>::iterator iter = find_if(v.begin(), v.end(), rnd);
v.insert(iter, rnd.getNum());
}
for (vector<int>::const_iterator iter = v.begin(); iter != v.end(); iter++)
cout << *iter << endl;
return 0;
}
答案 0 :(得分:2)
首先,您想要从空矢量开始:
std::vector<int> v;
其次,您当前的随机数生成器Random_int
仅在num
中生成一个号并保存。您需要的是Random_int
,其状态可以更新:
class random_int_state{
public:
random_int_state(int low, int high) :
re{/* Assignment 1 - use a suitable seed for the random generator.
* maybe you could use a `time`stamp or a `random device`? */
},
dist{low,high},
num(dist(re)) {}
int getNum() const
{
return num;
}
bool operator()(int x) const
{
return (num <= x);
}
void nextState(){
/* Assignment 2 - how do you create a new state?
** what is the observable state of this object after all? */
}
private:
std::mt19937 re;
std::uniform_int_distribution<int> dist;
int num;
};
除此之外,您的插入排序算法工作正常。只需记住在程序开头保持vector
为空,否则您将获得前导零。
请注意,此示例仍需要您的代码,但这很容易。
答案 1 :(得分:2)
#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <random>
int main() {
std::default_random_engine re { std::random_device()() };
std::uniform_int_distribution<int> dist;
std::vector<int> v;
for(int i=0; i!=10; ++i) {
int r=dist(re);
auto it=std::lower_bound(v.begin(), v.end(), r);
v.insert(it, r);
}
std::copy(
v.cbegin(), v.cend(),
std::ostream_iterator<int>(std::cout, "\n")
);
您的代码非常令人困惑,请考虑以上情况。因为向量是排序的,所以我们可以使用lower_bound
来找到插入O(log(n))而不是O(n)的正确位置