编写一个程序，以按排序的顺序打印给定字符串的所有不同排列。请注意，输入字符串可能包含重复的字符。
在数学中，置换的概念涉及将集合的所有成员按某种顺序或次序进行的动作，或者如果集合已经被排序，对其元素进行重新排列(重新排序)，则此过程称为置换。
资料来源–维基百科
例子：
Input : BAC
Output : ABC ACB BAC BCA CAB CBA
Input : AAB
Output : AAB ABA BAA
Input : DBCA
Output: ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA
使用的概念：由长度为’n’的不同字符的字符串生成的字符串数等于’n！’。对任何给定的字符串排序，然后从字典上生成下一个更大的字符串，直到从这些字符得出最大的字典上的字符串为止。
Different permutations of word “geeks”
Length of string = 5
Character ‘e’ repeats 2 times.
Result = 5!/2! = 60.
脚步：
示例：考虑字符串“ ABCD”。
步骤1：对字符串排序。
步骤2：获得可以由该字符串形成的排列总数。
步骤3：打印排序后的字符串，然后循环(排列1)次，因为已经打印了第一个字符串。
步骤4：找到下一个更大的字符串，。
这是此问题的实现–
C++
// C++ program to print all permutations
// of a string in sorted order.
#include
using namespace std;
// Calculating factorial of a number
int factorial(int n)
{
int f = 1;
for (int i = 1; i <= n; i++)
f = f * i;
return f;
}
// Method to find total number of permutations
int calculateTotal(string temp, int n)
{
int f = factorial(n);
// Building Map to store frequencies of
// all characters.
map hm;
for (int i = 0; i < temp.length(); i++)
{
hm[temp[i]]++;
}
// Traversing map and
// finding duplicate elements.
for (auto e : hm)
{
int x = e.second;
if (x > 1)
{
int temp5 = factorial(x);
f /= temp5;
}
return f;
}
}
static void nextPermutation(string &temp)
{
// Start traversing from the end and
// find position 'i-1' of the first character
// which is greater than its successor.
int i;
for (i = temp.length() - 1; i > 0; i--)
if (temp[i] > temp[i - 1]) break;
// Finding smallest character after 'i-1' and
// greater than temp[i-1]
int min = i;
int j, x = temp[i - 1];
for (j = i + 1; j < temp.length(); j++)
if ((temp[j] < temp[min]) and
(temp[j] > x))
min = j;
// Swapping the above found characters.
swap(temp[i - 1], temp[min]);
// Sort all digits from position next to 'i-1'
// to end of the string.
sort(temp.begin() + i, temp.end());
// Print the String
cout << temp << endl;
}
void printAllPermutations(string s)
{
// Sorting String
string temp(s);
sort(temp.begin(), temp.end());
// Print first permutation
cout << temp << endl;
// Finding the total permutations
int total = calculateTotal(temp, temp.length());
for (int i = 1; i < total; i++)
{
nextPermutation(temp);
}
}
// Driver Code
int main()
{
string s = "AAB";
printAllPermutations(s);
}
// This code is contributed by
// sanjeev2552 Java
// Java program to print all permutations of a string
// in sorted order.
import java.io.*;
import java.util.*;
class Solution {
// Calculating factorial of a number
static int factorial(int n) {
int f = 1;
for (int i = 1; i <= n; i++)
f = f * i;
return f;
}
// Method to print the array
static void print(char[] temp) {
for (int i = 0; i < temp.length; i++)
System.out.print(temp[i]);
System.out.println();
}
// Method to find total number of permutations
static int calculateTotal(char[] temp, int n) {
int f = factorial(n);
// Building HashMap to store frequencies of
// all characters.
HashMap hm =
new HashMap();
for (int i = 0; i < temp.length; i++) {
if (hm.containsKey(temp[i]))
hm.put(temp[i], hm.get(temp[i]) + 1);
else
hm.put(temp[i], 1);
}
// Traversing hashmap and finding duplicate elements.
for (Map.Entry e : hm.entrySet()) {
Integer x = (Integer)e.getValue();
if (x > 1) {
int temp5 = factorial(x);
f = f / temp5;
}
}
return f;
}
static void nextPermutation(char[] temp) {
// Start traversing from the end and
// find position 'i-1' of the first character
// which is greater than its successor.
int i;
for (i = temp.length - 1; i > 0; i--)
if (temp[i] > temp[i - 1])
break;
// Finding smallest character after 'i-1' and
// greater than temp[i-1]
int min = i;
int j, x = temp[i - 1];
for (j = i + 1; j < temp.length; j++)
if ((temp[j] < temp[min]) && (temp[j] > x))
min = j;
// Swapping the above found characters.
char temp_to_swap;
temp_to_swap = temp[i - 1];
temp[i - 1] = temp[min];
temp[min] = temp_to_swap;
// Sort all digits from position next to 'i-1'
// to end of the string.
Arrays.sort(temp, i, temp.length);
// Print the String
print(temp);
}
static void printAllPermutations(String s) {
// Sorting String
char temp[] = s.toCharArray();
Arrays.sort(temp);
// Print first permutation
print(temp);
// Finding the total permutations
int total = calculateTotal(temp, temp.length);
for (int i = 1; i < total; i++)
nextPermutation(temp);
}
// Driver Code
public static void main(String[] args) {
String s = "AAB";
printAllPermutations(s);
}
} Python3
# Python3 program to print
# all permutations of a
# string in sorted order.
from collections import defaultdict
# Calculating factorial
# of a number
def factorial(n):
f = 1
for i in range (1, n + 1):
f = f * i
return f
# Method to find total
# number of permutations
def calculateTotal(temp, n):
f = factorial(n)
# Building Map to store
# frequencies of all
# characters.
hm = defaultdict (int)
for i in range (len(temp)):
hm[temp[i]] += 1
# Traversing map and
# finding duplicate elements.
for e in hm:
x = hm[e]
if (x > 1):
temp5 = factorial(x)
f //= temp5
return f
def nextPermutation(temp):
# Start traversing from
# the end and find position
# 'i-1' of the first character
# which is greater than its successor
for i in range (len(temp) - 1, 0, -1):
if (temp[i] > temp[i - 1]):
break
# Finding smallest character
# after 'i-1' and greater
# than temp[i-1]
min = i
x = temp[i - 1]
for j in range (i + 1, len(temp)):
if ((temp[j] < temp[min]) and
(temp[j] > x)):
min = j
# Swapping the above
# found characters.
temp[i - 1], temp[min] = (temp[min],
temp[i - 1])
# Sort all digits from
# position next to 'i-1'
# to end of the string
temp[i:].sort()
# Print the String
print (''.join(temp))
def printAllPermutations(s):
# Sorting String
temp = list(s)
temp.sort()
# Print first permutation
print (''.join( temp))
# Finding the total permutations
total = calculateTotal(temp,
len(temp))
for i in range (1, total):
nextPermutation(temp)
# Driver Code
if __name__ == "__main__":
s = "AAB"
printAllPermutations(s)
# This code is contributed by ChitranayalC#
// C# program to print all permutations
// of a string in sorted order.
using System;
using System.Collections.Generic;
class GFG
{
// Calculating factorial of a number
static int factorial(int n)
{
int f = 1;
for (int i = 1; i <= n; i++)
f = f * i;
return f;
}
// Method to print the array
static void print(char[] temp)
{
for (int i = 0; i < temp.Length; i++)
Console.Write(temp[i]);
Console.WriteLine();
}
// Method to find total number of permutations
static int calculateTotal(char[] temp, int n)
{
int f = factorial(n);
// Building Dictionary to store frequencies
// of all characters.
Dictionary hm = new Dictionary();
for (int i = 0; i < temp.Length; i++)
{
if (hm.ContainsKey(temp[i]))
hm[temp[i]] = hm[temp[i]] + 1;
else
hm.Add(temp[i], 1);
}
// Traversing hashmap and
// finding duplicate elements.
foreach(KeyValuePair e in hm)
{
int x = e.Value;
if (x > 1)
{
int temp5 = factorial(x);
f = f / temp5;
}
}
return f;
}
static void nextPermutation(char[] temp)
{
// Start traversing from the end and
// find position 'i-1' of the first character
// which is greater than its successor.
int i;
for (i = temp.Length - 1; i > 0; i--)
if (temp[i] > temp[i - 1])
break;
// Finding smallest character after 'i-1'
// and greater than temp[i-1]
int min = i;
int j, x = temp[i - 1];
for (j = i + 1; j < temp.Length; j++)
if ((temp[j] < temp[min]) && (temp[j] > x))
min = j;
// Swapping the above found characters.
char temp_to_swap;
temp_to_swap = temp[i - 1];
temp[i - 1] = temp[min];
temp[min] = temp_to_swap;
// Sort all digits from position next to 'i-1'
// to end of the string.
Array.Sort(temp, i, temp.Length-i);
// Print the String
print(temp);
}
static void printAllPermutations(String s)
{
// Sorting String
char []temp = s.ToCharArray();
Array.Sort(temp);
// Print first permutation
print(temp);
// Finding the total permutations
int total = calculateTotal(temp, temp.Length);
for (int i = 1; i < total; i++)
nextPermutation(temp);
}
// Driver Code
public static void Main(String[] args)
{
String s = "AAB";
printAllPermutations(s);
}
}
// This code is contributed by Rajput-Ji 输出：
AAB
ABA
BAA
时间复杂度： O(n * m)，其中n是数组的大小，m是可能的排列数量。
辅助空间： O(n)。