打印给定两个字符串的所有交错

给定两个字符串str1 和 str2，编写一个函数来打印给定两个字符串的所有交错。您可以假设两个字符串中的所有字符都不同
例子：

Input: str1 = "AB",  str2 = "CD"
Output:
    ABCD
    ACBD
    ACDB
    CABD
    CADB
    CDAB

Input: str1 = "AB",  str2 = "C"
Output:
    ABC
    ACB
    CAB

给定两个字符串的交错字符串保留各个字符串中的字符顺序。例如，在上述第一个示例的所有交错中，“A”在“B”之前，“C”在“D”之前。

设 str1 的长度为 m，str2 的长度为 n。让我们假设 str1 和 str2 中的所有字符都是不同的。令 count(m, n) 为此类字符串中所有交错字符串的计数。 count(m, n) 的值可以写成如下。

count(m, n) = count(m-1, n) + count(m, n-1)
     count(1, 0) = 1 and count(0, 1) = 1

要打印所有交错，我们可以先修复输出字符串中 str1[0..m-1] 的第一个字符，然后递归调用 str1[1..m- 1 ] 和 str2[0..n-1]。然后我们可以修复str2[0..n-1]的第一个字符，递归调用str1[0..m-1]和str2[1..n- 1 ]。感谢 akash01 提供以下 C 实现。

C

// C program to print all interleavings of given two strings
#include 
#include 
#include 
 
// The main function that recursively prints all interleavings.
// The variable iStr is used to store all interleavings (or
// output strings) one by one. 
// i is used to pass next available place in iStr
void printIlsRecur (char *str1, char *str2, char *iStr, int m,
                    int n, int i)
{
    // Base case: If all characters of str1 and str2 have been
    // included in output string, then print the output string
    if (m==0 && n==0)
        printf("%s\n", iStr) ;
 
    // If some characters of str1 are left to be included, then
    // include the  first character from the remaining characters
    // and recur for rest
    if (m != 0)
    {
        iStr[i] = str1[0];
        printIlsRecur (str1 + 1, str2, iStr, m-1, n, i+1);
    }
 
    // If some characters of str2 are left to be included, then
    // include the  first character from the remaining characters
    // and recur for rest
    if (n != 0)
    {
        iStr[i] = str2[0];
        printIlsRecur(str1, str2+1, iStr, m, n-1, i+1);
    }
}
 
// Allocates memory for output string and uses printIlsRecur()
// for printing all interleavings
void printIls (char *str1, char *str2, int m, int n)
{
   // allocate memory for the output string
   char *iStr= (char*)malloc((m+n+1)*sizeof(char));
 
   // Set the terminator for the output string
   iStr[m+n] = '\0';
 
   // print all interleavings using printIlsRecur()
   printIlsRecur (str1, str2, iStr, m, n, 0);
 
   // free memory to avoid memory leak
   free(iStr);
}
 
// Driver program to test above functions
int main()
{
    char str1[] = "AB";
    char str2[] = "CD";
    printIls (str1, str2, strlen(str1), strlen(str2));
    return 0;
}

C++

// C++ program to print all interleavings of given two strings
#include 
using namespace std;
 
// The main function that recursively prints all interleavings.
// The variable iStr is used to store all interleavings (or
// output strings) one by one.
// i is used to pass next available place in iStr
void printIlsRecur (char *str1, char *str2, char *iStr, int m,
                    int n, int i)
{
    // Base case: If all characters of str1 and str2 have been
    // included in output string, then print the output string
    if (m == 0 && n == 0)
        cout << iStr << endl ;
 
    // If some characters of str1 are left to be included, then
    // include the first character from the remaining characters
    // and recur for rest
    if (m != 0)
    {
        iStr[i] = str1[0];
        printIlsRecur (str1 + 1, str2, iStr, m - 1, n, i + 1);
    }
 
    // If some characters of str2 are left to be included, then
    // include the first character from the remaining characters
    // and recur for rest
    if (n != 0)
    {
        iStr[i] = str2[0];
        printIlsRecur(str1, str2 + 1, iStr, m, n - 1, i + 1);
    }
}
 
// Allocates memory for output string and uses printIlsRecur()
// for printing all interleavings
void printIls (char *str1, char *str2, int m, int n)
{
    // allocate memory for the output string
    char *iStr= new char[((m + n + 1)*sizeof(char))];
     
    // Set the terminator for the output string
    iStr[m + n] = '\0';
     
    // print all interleavings using printIlsRecur()
    printIlsRecur (str1, str2, iStr, m, n, 0);
     
    // free memory to avoid memory leak
    free(iStr);
}
 
// Driver code
int main()
{
    char str1[] = "AB";
    char str2[] = "CD";
    printIls (str1, str2, strlen(str1), strlen(str2));
    return 0;
}
 
// This is code is contributed by rathbhupendra

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    /*
     * This methods prints interleaving string of two
     * strings
     * @param s1  String 1
     * @param i   current index of s1
     * @param s2  String 2
     * @param j  Current index of s2
     * @param asf String containing interleaving string of
     *     s1 and s2
     *
     */
    static void printInterLeaving(String s1, int i,
                                  String s2, int j,
                                  String asf)
    {
        if (i == s1.length() && j == s2.length()) {
            System.out.println(asf);
        }
 
        // Either we will start with string 1
        if (i < s1.length())
            printInterLeaving(s1, i + 1, s2, j,
                              asf + s1.charAt(i));
        // Or with string 2
        if (j < s2.length())
            printInterLeaving(s1, i, s2, j + 1,
                              asf + s2.charAt(j));
    }
 
    /*
     * Main function executed by JVM
     * @param args String array
     */
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub
 
        String s1 = "AB"; // String 1
        String s2 = "CD"; // String 2
       
        printInterLeaving(s1, 0, s2, 0, "");
    }
}
 
/* Code by mahi_07 */

Python3

# Python program to print all interleavings of given two strings
 
# Utility function
def toString(List):
    return "".join(List)
 
# The main function that recursively prints all interleavings.
# The variable iStr is used to store all interleavings (or output
# strings) one by one.
# i is used to pass next available place in iStr
def printIlsRecur(str1, str2, iStr, m, n, i):
 
    # Base case: If all characters of str1 and str2 have been
    # included in output string, then print the output string
    if m==0 and n==0:
        print (toString(iStr))
 
    # If some characters of str1 are left to be included, then
    # include the first character from the remaining characters
    # and recur for rest
    if m != 0:
        iStr[i] = str1[0]
        printIlsRecur(str1[1:], str2, iStr, m-1, n, i+1)
 
    # If some characters of str2 are left to be included, then
    # include the first character from the remaining characters
    # and recur for rest
    if n != 0:
        iStr[i] = str2[0]
        printIlsRecur(str1, str2[1:], iStr, m, n-1, i+1)
 
# Allocates memory for output string and uses printIlsRecur()
# for printing all interleavings
def printIls(str1, str2, m, n):
    iStr = [''] * (m+n)
 
    # print all interleavings using printIlsRecur()
    printIlsRecur(str1, str2, iStr, m, n, 0)
 
# Driver program to test the above function
str1 = "AB"
str2 = "CD"
printIls(str1, str2, len(str1), len(str2))
 
# This code is contributed by Bhavya Jain

输出：

ABCD
ACBD
ACDB
CABD
CADB
CDAB

时间复杂度： O(2 ^ (m+n))