编写一个递归函数来反转输入字符串

Question

I've been reading the book C++ For Everyone and one of the exercises said to write a function string reverse(string str) where the return value is the reverse of str .

Can somebody write some basic code and explain it to me? I've been staring at this question since yesterday and can't figure it out. The furthest I've gotten is having the function return the first letter of str (Which I still don't know how it happened)

This is as far as I got (An hour after posting this question):

string reverse(string str)
{
    string word = "";

    if (str.length() <= 1)
    {
        return str;
    }
    else
    {
        string str_copy = str;
        int n = str_copy.length() - 1;
        string last_letter = str_copy.substr(n, 1);

        str_copy = str_copy.substr(0, n);
        word += reverse(str_copy);
        return str_copy;
    }
    return word;
}

If I enter "Wolf", it returns Wol. Somebody help me out here If I return word instead of return str_copy then I get a w If I return last_letter then I get an l

Answer 1

I'll instead explain the recursive algorithm itself. Take the example "input" which should produce "tupni". You can reverse the string recursively by

• If the string is empty or a single character, return it unchanged.
• Otherwise,
  1. Remove the first character.
  2. Reverse the remaining string.
  3. Add the first character above to the reversed string.
  4. Return the new string.

Answer 2

Try this one

string reverse(string &s)
{
    if( s.length() == 0 )  // end condtion to stop recursion
        return "";

    string last(1,s[s.length()-1]);  // create string with last character
    string reversed = reverse(s.substr(0,s.length()-1));
    return last+reversed; // Make he last character first
}

A recursive function must have the following properties

• It must call itself again
• It must have a condition when the recursion ends. Otherwise you have a function which will cause a stack overflow.

This recursive function does basically create a string of the last character and then call itself again with the rest of the string excluding the last character. The real switching happens at the last line where last+reversed is returned. If it would be the other way around nothing would happen.

It is very inefficient but it works to show the concept.

Answer 3

Just to suggest a better way of handling recursion:

String reversal using recursion in C++:

#include <iostream>
#include <string>
using namespace std;

string reverseStringRecursively(string str){
    if (str.length() == 1) {
        return str;
    }else{
        return reverseStringRecursively(str.substr(1,str.length())) + str.at(0);
    }
}

int main()
{
    string str;
    cout<<"Enter the string to reverse : ";
    cin>>str;

    cout<<"The reversed string is : "<<reverseStringRecursively(str);
    return 0;
}

Answer 4

Here is my version of a recursive function that reverses the input string:

void reverse(char *s, size_t len)
{
    if ( len <= 1 || !s )
    {
        return;
    }
    std::swap(s[0], s[len-1]);// swap first and last simbols
    s++; // move pointer to the following char
    reverse(s, len-2); // shorten len of string
}

Answer 5

I won't write a full-blown algorithm for you, but here's a hint:

How about swapping the two outermost characters, and then apply the same to the characters in the middle?

Oh, and if that book really proposed string reverse(string str) as an appropriate function signature for this, throw it away and buy a good book instead.

Answer 6

Shortest and easiest

class Solution {
public:
    string reverseString(string s) {
        string str;
        if(s != "\0"){
            str = reverseString(s.substr(1, s.length()));
            str += s.substr(0,1);
        }
        return str;    
    }   
};

Answer 7

I know I shouldn't give a solution, but since no one mentioned this easy solution I though I should share it. I think the code literally is the algorithm so there is no need for a pseudo-code.

void c_plusplus_recursive_swap_reverse(std::string::iterator start, 
    std::string::iterator end) 
{
    if(start >= end) {
        return;
    }

    std::iter_swap(start, end);
    c_plusplus_recursive_swap_reverse(++start, --end);
}

To call it use:

c_plusplus_recursive_swap_reverse(temp.begin(), temp.end());

Answer 8

1-line recursive solution:

string RecursiveReverse(string str, string prev = "") {
    return (str.length() == 0 ? prev : RecursiveReverse(str.substr(0, str.length()-1), prev += str[str.length()-1]));
}

You call it like this:

cout << RecursiveReverse("String to Reverse");

Answer 9

All existing solutions had way too much code that didn't really do anything, so, here's my take at it:

#include <iostream>
#include <string>

std::string
r(std::string s)
{
    if (s.empty())
        return s;
    return r(s.substr(1)) + s[0];
}

int
main()
{
    std::cout << r("testing") << std::endl;
}

---

PS I stumbled upon this question trying to find a C++ way for std::string of what s+1 for a char * in C is; without going the whole route of s.substr(1, s.length()-1) , which looks too ugly. Turns out, there's std::string::npos , which means until the end of the string, and it's already the default value for the second argument, so, s.substr(1) is enough (plus, it also looks more efficient and on par with the simple s + 1 in C).

---

Note, however, that recursion in general doesn't scale as the input grows larger, unless the compiler is able to do what is known as tail-recursion optimisation. (Recursion is rarely relied upon in imperative languages.)

However, in order for the tail recursion optimisation to get activated, it is generally required that, (0), the recursion only happens within the return statement, and that, (1), no further operations are performed with the result of the recursive call back in the parent function.

Eg, in the case above, the + s[0] is logically done by the parent after the child call completes (and it probably would be so even if you go the more uglier s[s.length()-1] + route), so, it might as well prevent most compilers from doing a tail-recursion-optimisation, thus making the function very inefficient on large inputs (if not outright broken due to heap exhaustion).

(For what it's worth, I've tried writing a more tail-recursion-friendly solution (making sure to grow the return result through an argument to the function itself), but disassembly of the resulting binary seems to suggest that it's more involved than that in the imperative languages like C++, see gcc: is there no tail recursion if I return std::string in C++? .)

Answer 10

you can implement your own reverse similar to std::reverse.

template <typename BidirIt>
void reverse(BidirIt first, BidirIt last)
{
    if((first == last) || (first == --last))
        return;

    std::iter_swap(first, last);
    reverse(++first, last);
}

Answer 11

I did something like this, it did the reversal in place. I took two variables that traverse the string from two extreme end to the centre of the string and when they overlap or equal to each other then reversal terminates.

Take an example: input string str = "abcd" and call the function as

ReverseString(str,0,str.length()-1);

and increment/decrement the variable pointers recursively. First the pointers points to 'a' and 'd' and swap them, then they point to 'b' and 'c' and swap them. Eventually i >= j which calls for the base case to be true and hence the recursion terminates. The main take away for this question is to pass input string as reference.

string ReverseString(string& str,int i,int j){
        if(str.length() < 1 || str == "" || i >= j){
            return "";
        }

        else{
            char temp = str[i];
            str[i] = str[j];
            str[j] = temp;
            ReverseString(str,i+1,j-1);
        }
        return str;
    }

Answer 12

String can be reversed in-place. If we start from smallest possible string ie one character string, we don't need to do anything. This is where we stop or return from our recursive call and it becomes our base case.

Next, we have to think of a generic way to swap the smallest string ie two characters or more. Simplest logic is to swap the current character str[current_index] with character on the opposite side str[str_length-1 - current_index] .

In the end, call the reverse function again for next index.

#include <iostream>
using namespace std;

void reverse_string(std::string& str, int index, int length) {
  // Base case: if its a single element, no need to swap
  // stop swapping as soon as we reach the mid, hence index*2
  // otherwise we will reverse the already reversed string
  if( (length - index*2) <= 1 ) { 
    return;
  }

  // Reverse logic and recursion:

  // swap current and opposite index
  std::swap(str[index], str[length-1 - index]); 

  // do the same for next character (index+1)
  reverse_string(str, index+1, length);
}

int main() {
  std::string s = "World";
  reverse_string(s, 0, s.length());
  std::cout << s << endl;
}

Answer 13

There are already some good answer but I want to add my approach with full working Recursive reversing string.

#include <iostream>
#include <string>
using namespace std;

char * reverse_s(char *, char*, int,int);

int main(int argc, char** argv) {
if(argc != 2) {
        cout << "\n ERROR! Input String";
        cout << "\n\t " << argv[0] << "STRING" << endl;
        return 1;
}       
        char* str = new char[strlen(argv[1])+1];
        strcpy(str,argv[1]);    
        char* rev_str = new char[strlen(str)+1];        
        cout<<"\n\nFinal Reverse of '" << str << "' is --> "<< reverse_s(str, rev_str, 0, strlen(str)) << endl;
        cin.ignore();
        delete rev_str, str;
        return 0;
}

char* reverse_s(char* str, char* rev_str, int str_index, int rev_index ) {
if(strlen(str) == 1)
        return str;

if(str[str_index] == '\0' ) {
        rev_str[str_index] = '\0';
        return rev_str;
}

str_index += 1;
rev_index -=1;

rev_str = reverse_s(str, rev_str, str_index, rev_index);
if(rev_index >= 0) {
        cout << "\n Now the str value is " << str[str_index-1] << " -- Index " << str_in
dex << " Rev Index: " << rev_index;
        rev_str[rev_index] = str[str_index-1];

        cout << "\nReversed Value: " << rev_str << endl;
}
return rev_str;
}

Answer 14

void reverse(string &s, int &m) {
    if (m == s.size()-1)
        return;
    int going_to = s.size() - 1 - m;
    string leader = s.substr(1,going_to);
    string rest = s.substr(going_to+1,s.size());
    s = leader + s.substr(0,1) + rest;
    reverse(s,++m);    
}
int main ()
{
  string y = "oprah";
  int sz = 0;
  reverse(y,sz);
  cout << y << endl;
  return 0;
}

Answer 15

here is my 3 line string revers

std::string stringRevers(std::string s)
{
    if(s.length()<=1)return s;
    string word=s.at(s.length()-1)+stringRevers(s.substr(0,s.length()-1));//copy the last one at the beginning  and do the same with the rest
    return word;

}

Answer 16

void ClassName::strgRevese(char *str)
{
        if (*str=='\0')
                return;
        else
                strgRevese(str+1);
        cout <<*str;
}

Answer 17

The question is to write a recursive function. Here is one approach. Not a neat code, but does what is required.

/* string reversal through recursion */
#include <stdio.h>
#include <string.h>
#define size 1000
char rev(char []);
char new_line[size];
int j = 0;
int i =0;
int main ()
{
  char string[]="Game On";
  rev(string);
  printf("Reversed rev string is %s\n",new_line);
  return 0;
}
char rev(char line[])
{
 while(line[i]!='\0')
  { 
    i++;
    rev(line);
    i--;
    new_line[j] = line[i];
    j++;
    return line[i];
  }
  return line[i];
}

Answer 18

It will reverse Original string recursively

void swap(string &str1, string &str2)
{
    string temp = str1;
    str1 = str2;
    str2 = str1;
}

void ReverseOriginalString(string &str, int p, int sizeOfStr)
{
    static int i = 0;
    if (p == sizeOfStr)
        return;

    ReverseOriginalString(str, s + 1, sizeOfStr);

    if (i <= p)
        swap(&str[i++], &str[p])
}

int main()
{
    string st = "Rizwan Haider";

    ReverseOriginalString(st, 0, st.length());
    std::cout << "Original String is Reversed: " << st << std::endl;

    return 0;
}