编程实现:string reverse(string &s),用递归算法使字符串s倒序

#include<iostream>

#include<string>

using namespace std;

string reverse(string s){

static string a;

int t1=s.length(),t2=a.length();

if(t1==t2)

{s=a;

return s;}

else a+=s.at(t1-t2-1);

reverse(s);

}

int main() {

string str;

cin >> str;

cout << "原字符串为:" << str << endl;

cout << "倒序反转后为:" << reverse(str) << endl;

return 0;

}



为什么主函数第二个cout没有输出?是我的逻辑出现错误了吗。

修改如下:
只要在 reverse(s) 改为 return reverse(s); 就好了

#include<iostream>
 
#include<string>
 
using namespace std;
 
string reverse(string s){
 
static string a;
 
int t1=s.length(),t2=a.length();
 
if(t1==t2)
 
{s=a;
 
return s;}
 
else a+=s.at(t1-t2-1);
 
return reverse(s);
 
}
 
int main() {
 
string str;
 
cin >> str;
 
cout << "原字符串为:" << str << endl;
 
cout << "倒序反转后为:" << reverse(str) << endl;
 
return 0;
 
}
 
 
 

应该是void类型的返回值才能实现吧

img

#include<iostream>
#include<string>
using namespace std;
void reverse(string &s)
{
    static string a;
    int t1=s.length(),t2=a.length();
    if(t1==t2)
    {
        s=a;
        return;
    }
    else
        a+=s.at(t1-t2-1);
    reverse(s);

}
int main()
{
    string str;
    cin >> str;
    cout << "原字符串为:" << str << endl;
   reverse(str);
    cout << "倒序反转后为:" << str << endl;
    return 0;
}





string  Resvers(string str) {
    int len = str.size();
    if (len == 1)
        return str;
    if (len == 2) {
        swap(str[0], str[1]);
        return str;
    }
    auto cs = str[0];
    auto ce = str[len - 1];
    return ce + Resvers(str.substr(1, len - 2))+cs;
}
int main()
{
    string str;
    cin >> str;
    cout << Resvers(str);
}
您好,我是有问必答小助手,您的问题已经有小伙伴帮您解答,感谢您对有问必答的支持与关注!
PS:问答VIP年卡 【限时加赠:IT技术图书免费领】,了解详情>>> https://vip.csdn.net/askvip?utm_source=1146287632