[开发工具]c++递归学习

字符反转递归

#include<bits/stdc++.h>
using namespace std;
int sum(int a[],int start){
	if(start==9) return a[start];
	return a[start]+sum(a,start+1);
}
string fan(string s,int end){
	if(end<0) return "";
	return fan(s.substr(1),end-1)+s[0];
}
int fibo(int n){
	if(n==1||n==2)
	return 1;
	return fibo(n-1)+fibo(n-2);
}
int main()
{
	int a[10]={0};
	for(int i=1;i<10;i++)
	a[i]=i;
	cout<<sum(a,0)<<endl;	
	string s="iloveyou";
	cout<<fan(s,s.length()-1)<<endl;
	cout<<fibo(3)<<endl;
	return 0;
}

fibo递归andfor循环

#include<bits/stdc++.h>
using namespace std;
int fibo(int n)
{
	if(n==1||n==2)
	return 1;
	return fibo(n-1)+fibo(n-2);
}
int n[10000];
int main()
{	
	n[1]=n[2]=1;	
	int a;
	cin>>a;
	cout<<fibo(a)<<endl;
	for(int i=3;i<=a;i++)
	{
		n[i]=n[i-1]+n[i-2];
	}
	cout<<n[a]<<endl;
	return 0;
}

汉诺塔

#include<bits/stdc++.h>
using namespace std;
void HanoiTower(int N,string from,string to,string help){
	if(N==1){
		cout<<"move"<<N<<from<<"->"<<to<<endl;
		return;
	}
	HanoiTower(N-1,from,help,to);
	cout<<"move"<<N<<from<<"->"<<to<<endl;
	HanoiTower(N-1,help,to,from);
}
int main()
{
	int a;
	string from,to,help;
	cin>>a>>from>>to>>help;
	HanoiTower(a,from,to,help);
	return 0;
}

二分查找递归and非递归

#include<bits/stdc++.h>
using namespace std;
int find(int n[],int lefe,int right,int temp){//lower_bound;
	if(lefe>=right){
	if(lefe==right)
	return lefe;		
		return -1;
	}
	int mid=(lefe+right)>>1;
	if(n[mid]>=temp){
		find(n,lefe,mid,temp);
	}
	else if(n[mid]<temp){
		find(n,mid+1,right,temp);
	}
}
int find1(int n[],int lefe,int right,int temp){//upper_bound	
	if(lefe>=right){
	if(lefe==right)
	return lefe;		
		return -1;
	}		
	int mid=(right+lefe+1)>>1;
	if(n[mid]<=temp){
		find1(n,mid,right,temp);
	}
	else if(n[mid]>temp){
		find1(n,lefe,mid-1,temp);
	}
	else return lefe;
	
}
int main()
{
	srand((unsigned)time(NULL));
	int n[110];
	for(int i=0;i<110;i++){
		n[i]=i+rand()%10+1;
	}
	sort(n,n+110);
	for(int i=0;i<8;i++){
		cout<<n[i]<<" ";
	}
	cout<<endl;
	int temp;
	cin>>temp;
	cout<<find(n,0,109,temp)<<endl;
	cout<<find1(n,0,109,temp)<<endl;
	int l=0,r=109;
	while(l<r){
		int mid=(l+r)>>1;
		if(n[mid]>=temp){
			r=mid;
		}
		else
		    l=mid+1;    
	}
	cout<<l<<endl;
	l=0,r=109;
	while(l<r){
		int mid=(l+r+1)>>1;
		if(n[mid]<=temp){
			l=mid;
		}
		else
		    r=mid-1;    
	}
	cout<<l<<endl;
	return 0;
}

最小公倍数最大公约数

#include<iostream>
using namespace std;
typedef long long LL;
LL lcm(LL a,LL b)
{
	return a%b==0?b:lcm(b,a%b);
}
int main()
{
	LL a,b,c;
	cin>>a>>b;
	cout<<lcm(a,b)<<endl; 
	LL temp=a*b/lcm(a,b);
	cout<<temp<<endl;
	return 0;
}

总结：大问题转换小问题，不改变问题本质，找边界，找变化。