洗牌算法——N個數隨機打亂

 具體思路：

1.隨機生成一個(1,N)的整數，代表第一個數的數組下標

2.用erase刪除vector容器中該數，然後再隨機生成(1,N-1)，代表第二個數的數組下標

3.遍歷N-1次

代碼如下：

#include <iostream>
#include <vector>
#include <ctime> 
#include<random>
using namespace std;
class RandomNumber
{
public:
	RandomNumber() {
		srand((unsigned)time(NULL));    //析構函數，在對象創建時數據成員執行初始化操作
	}
	int integer(int begin, int end)
	{
		return rand() % (end - begin + 1) + begin;
	}
};
vector<double> random_shuffle(vector<double>a) 
{
	extern RandomNumber r;
	int n = size(a), x;
	vector<double>b;
	for (int i = 0; i < n - 1; i++)
	{
		x = r.integer(0, n-1-i);
		//cout << x << endl;
		b.push_back(a[x]);
		for (auto ite = a.begin(); ite != a.end();) {
			if (*ite == b[i])
				ite = a.erase(ite);
			else
			{
				++ite;
			}
		}
	}
	b.push_back(a[0]);
	return b;
}

RandomNumber r;      
int main()
{
	vector<double>a;
	cout << "before random shuffle:";
	for (int i = 0; i < 10; i++)
	{
		a.push_back(i);
		cout << a[i];
	}cout << endl;
	vector<double>b;
	for (int i = 0; i < 10; i++)
	{
	b=random_shuffle(a);
	cout << "after random shuffle  "<<i<<":";
	for (int i = 0; i < 10; i++)
	{
		cout << b[i];
	}cout << endl;
	}
}

輸出結果：