编写一个程序来打印给定字符串的所有排列

排列，也称为“排列编号”或“顺序”，是将有序列表 S 的元素重新排列为与 S 本身一一对应的排列。长度为 n 的字符串有 n!排列。

来源：Mathword(http://mathworld.wolfram.com/Permutation.html)

下面是字符串ABC 的排列。
ABC ACB BAC BCA CBA CAB

这是一个用作回溯基础的解决方案。

C++

// C++ program to print all
// permutations with duplicates allowed
#include 
using namespace std;
 
 
// Function to print permutations of string
// This function takes three parameters:
// 1. String
// 2. Starting index of the string
// 3. Ending index of the string.
void permute(string a, int l, int r)
{
    // Base case
    if (l == r)
        cout<

---

C

// C program to print all permutations with duplicates allowed
#include 
#include 
 
/* Function to swap values at two pointers */
void swap(char *x, char *y)
{
    char temp;
    temp = *x;
    *x = *y;
    *y = temp;
}
 
/* Function to print permutations of string
This function takes three parameters:
1. String
2. Starting index of the string
3. Ending index of the string. */
void permute(char *a, int l, int r)
{
int i;
if (l == r)
    printf("%s\n", a);
else
{
    for (i = l; i <= r; i++)
    {
        swap((a+l), (a+i));
        permute(a, l+1, r);
        swap((a+l), (a+i)); //backtrack
    }
}
}
 
/* Driver program to test above functions */
int main()
{
    char str[] = "ABC";
    int n = strlen(str);
    permute(str, 0, n-1);
    return 0;
}

Java

// Java program to print all permutations of a
// given string.
public class Permutation
{
    public static void main(String[] args)
    {
        String str = "ABC";
        int n = str.length();
        Permutation permutation = new Permutation();
        permutation.permute(str, 0, n-1);
    }
 
    /**
    * permutation function
    * @param str string to calculate permutation for
    * @param l starting index
    * @param r end index
    */
    private void permute(String str, int l, int r)
    {
        if (l == r)
            System.out.println(str);
        else
        {
            for (int i = l; i <= r; i++)
            {
                str = swap(str,l,i);
                permute(str, l+1, r);
                str = swap(str,l,i);
            }
        }
    }
 
    /**
    * Swap Characters at position
    * @param a string value
    * @param i position 1
    * @param j position 2
    * @return swapped string
    */
    public String swap(String a, int i, int j)
    {
        char temp;
        char[] charArray = a.toCharArray();
        temp = charArray[i] ;
        charArray[i] = charArray[j];
        charArray[j] = temp;
        return String.valueOf(charArray);
    }
 
}
 
// This code is contributed by Mihir Joshi

Python

# Python program to print all permutations with
# duplicates allowed
 
def toString(List):
    return ''.join(List)
 
# Function to print permutations of string
# This function takes three parameters:
# 1. String
# 2. Starting index of the string
# 3. Ending index of the string.
def permute(a, l, r):
    if l==r:
        print toString(a)
    else:
        for i in xrange(l,r+1):
            a[l], a[i] = a[i], a[l]
            permute(a, l+1, r)
            a[l], a[i] = a[i], a[l] # backtrack
 
# Driver program to test the above function
string = "ABC"
n = len(string)
a = list(string)
permute(a, 0, n-1)
 
# This code is contributed by Bhavya Jain

C#

// C# program to print all
// permutations of a given string.
using System;
 
class GFG
{
    /**
    * permutation function
    * @param str string to
    calculate permutation for
    * @param l starting index
    * @param r end index
    */
    private static void permute(String str,
                                int l, int r)
    {
        if (l == r)
            Console.WriteLine(str);
        else
        {
            for (int i = l; i <= r; i++)
            {
                str = swap(str, l, i);
                permute(str, l + 1, r);
                str = swap(str, l, i);
            }
        }
    }
 
    /**
    * Swap Characters at position
    * @param a string value
    * @param i position 1
    * @param j position 2
    * @return swapped string
    */
    public static String swap(String a,
                            int i, int j)
    {
        char temp;
        char[] charArray = a.ToCharArray();
        temp = charArray[i] ;
        charArray[i] = charArray[j];
        charArray[j] = temp;
        string s = new string(charArray);
        return s;
    }
 
// Driver Code
public static void Main()
{
    String str = "ABC";
    int n = str.Length;
    permute(str, 0, n-1);
}
}
 
// This code is contributed by mits

Javascript

// Javascript program to print all permutations of a
// given string.
 
function permute(str, l, r)
{
    if (l == r)
            document.write(str+"<br>");
        else
        {
            for (let i = l; i <= r; i++)
            {
                str = swap(str, l, i);
                permute(str, l + 1, r);
                str = swap(str, l, i);
            }
        }
}
 
function swap(a, i, j)
{
    let temp;
let charArray = a.split("");
temp = charArray[i] ;
charArray[i] = charArray[j];
charArray[j] = temp;
return (charArray).join("");
}
 
let str = "ABC";
let n = str.length;
permute(str, 0, n-1);
 
// This code is contributed by avanitrachhadiya2155

输出：

ABC
ACB
BAC
BCA
CBA
CAB

算法范式：回溯

时间复杂度： O(n*n!) 注意有 n!排列，打印排列需要 O(n) 时间。

注意：如果输入字符串有重复字符，上述解决方案将打印重复排列。请参阅下面的链接以获取即使输入中有重复也只打印不同排列的解决方案。
打印具有重复项的给定字符串的所有不同排列。
使用 STL 对给定字符串进行排列

另一种方法：

C++

#include 
#include 
using namespace std;
 
void permute(string s , string answer)
{
    if(s.length() == 0)
    {
        cout<>s;
 
    cout<<"\nAll possible strings are : ";
    permute(s , answer);
    return 0;
}

Java

import java.util.*;
 
class GFG{
  
static void permute(String s , String answer)
{  
    if (s.length() == 0)
    {
        System.out.print(answer + "  ");
        return;
    }
     
    for(int i = 0 ;i < s.length(); i++)
    {
        char ch = s.charAt(i);
        String left_substr = s.substring(0, i);
        String right_substr = s.substring(i + 1);
        String rest = left_substr + right_substr;
        permute(rest, answer + ch);
    }
}
 
// Driver code
public static void main(String args[])
{
    Scanner scan = new Scanner(System.in);
     
    String s;
    String answer="";
     
    System.out.print("Enter the string : ");
    s = scan.next();
     
    System.out.print("\nAll possible strings are : ");
    permute(s, answer);
}
}
 
// This code is contributed by adityapande88

蟒蛇3

def permute(s, answer):
    if (len(s) == 0):
        print(answer, end = "  ")
        return
     
    for i in range(len(s)):
        ch = s[i]
        left_substr = s[0:i]
        right_substr = s[i + 1:]
        rest = left_substr + right_substr
        permute(rest, answer + ch)
 
# Driver Code
answer = ""
 
s = input("Enter the string : ")
 
print("All possible strings are : ")
permute(s, answer)
 
# This code is contributed by Harshit Srivastava

Output:
Enter the string : abc
All possible strings are : abc  acb  bac  bca  cab  cba

时间复杂度： O(n*n!) 时间复杂度和上面的方法一样，即有n！排列，打印排列需要 O(n) 时间。

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