回溯解决爬楼梯问题

Posted 2020-09-24 ljbguanli

1.问题描写叙述

每次爬楼梯有每次可跨1,2,3步。爬上一个N阶楼梯有多少种方式，打印出每种方式。

2.源码

// ConsoleApplication6.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include <stdio.h>
#include <vector>
#include <map>
using namespace std;

typedef std::vector<int> VecStep;
typedef std::map<int, VecStep> MapWay;

VecStep m_vecCanStep;
VecStep m_vecNextStep;
MapWay m_mapWay;
int iSum = 0;

void wayPrint()
{
	for (auto it = m_mapWay.begin(); it != m_mapWay.end(); it++)
	{
		int iSize = it->second.size();
		for (int i = 0; i < iSize; i++)
		{
			printf("%d ", it->second[i]);
		}
		printf("\n");
	}
}

void creep(int iLeftStep, int iNextStep, int iCnt)
{
	for (int i = 0; i < (int)m_vecCanStep.size(); i++)
	{
		if (iNextStep == m_vecCanStep[i])
		{
			int iStepSize = m_vecNextStep.size();
			if (iCnt >= iStepSize)
			{
				m_vecNextStep.push_back(iNextStep);
			}
			else if (iCnt < iStepSize)
			{
				m_vecNextStep[iCnt] = iNextStep;
			}
			iCnt++;
		}
	}
	if (iLeftStep == 0)
	{
		VecStep vecStep;
		for (int i = 0; i < iCnt; i++)
		{
			vecStep.push_back(m_vecNextStep[i]);
		}
		m_mapWay.insert(MapWay::value_type(iSum, vecStep));
		iSum++;
	}

	for (int i = 0; i < (int)m_vecCanStep.size(); i++)
	{
		if (iLeftStep - m_vecCanStep[i] >= 0)
		{
			creep(iLeftStep - m_vecCanStep[i], m_vecCanStep[i], iCnt);
		}
	}
}

void main()
{
	int iTotalStep = 0;
	int iNextStep = 0;                              
	int iStepCnt = 0;                           
	m_vecCanStep.push_back(1);
	m_vecCanStep.push_back(2);
	m_vecCanStep.push_back(3);

	scanf_s("%d", &iTotalStep);

	creep(iTotalStep, iNextStep,iStepCnt);
	wayPrint();
	printf("TotalWay = %d", iSum);
}

3.代码解析

m_vecCanStep：每次能够走的步数，插入1,2,3

m_vecNextStep：从0-iCnt为每种能够达到的N阶的走法的步骤

m_mapWay：全部的走法

Creep形參含义：iLeftStep剩余总阶数，iNextStep下一步走法（1,2,3），iCnt当前走法已经走过的步骤统计

4.执行结果