DFS 对一系列给定数据 进行深度优先遍历 获得54321的所有可能组合

深度优先遍历，获得54321的所有可能组合

// DFS.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "stack"
#include "stdio.h"
#include "iostream"
#include "vector"

using namespace std;
vector<int> vec;
vector<int> vec2;
stack<int> s;

void DFS(int i);
void DFS2();

int _tmain(int argc, _TCHAR* argv[])
{
	//DFS(0);
	//vec[0] = 5;
	for (int i = 0;i<5;i++)
	{
		vec.push_back(i+1);
	}
	DFS2();
	system("pause");
	return 0;
}

void DFS2()
{
	if (vec2.size() == 5)
	{
		for (int i = 0; i<vec2.size();i++)
		{
			printf("%4d", vec2[i]);
		}
		printf("\n");
	}
	else
	{
		for (int i=0;i<vec.size();i++)
		{
			int t = vec[i];
			s.push(t);
			vec2.push_back(t);
			vec.erase(vec.begin()+i);
			DFS2();
			s.pop();
			vec2.pop_back();
			vec.insert(vec.begin()+i,t);
		}
	}
}

void DFS(int i)
{
	if (i==5)
	{
		for (int i = 0;i<vec.size();i++)
		{
			printf("%4d",vec[i]);
		}
		printf("\n");
	}

	if (i<5)
	{
		s.push(5);vec.push_back(5);
		DFS(i+1);
		s.pop();vec.pop_back();
	}

	if (i<5)
	{
		s.push(4);vec.push_back(4);
		DFS(i+1);
		s.pop();vec.pop_back();
	}

	if (i<5)
	{
		s.push(3);vec.push_back(3);
		DFS(i+1);
		s.pop();vec.pop_back();
	}

	if (i<5)
	{
		s.push(2);vec.push_back(2);
		DFS(i+1);
		s.pop();vec.pop_back();
	}

	if (i<5)
	{
		s.push(1);vec.push_back(1);
		DFS(i+1);
		s.pop();vec.pop_back();
	}
}