“中国浙江杭州”这样的一串字符串有多少个不重复的排序组合？要求：长度不变，不能重复。
1) java实现：
import java.util.ArrayList;
import java.util.List;
public class StringCombination {
public static void main(String[] args) {
String input = "中国浙江杭州";
int length = 6;
List<String> combinations = generateCombinations(input, length);
System.out.println("满足条件的组合数量：" + combinations.size());
System.out.println("满足条件的组合：");
for (String combination : combinations) {
System.out.println(combination);
}
}
private static List<String> generateCombinations(String input, int length) {
List<String> combinations = new ArrayList<>();
generateCombinationsHelper(input, length, "", combinations);
return combinations;
}
private static void generateCombinationsHelper(String input, int length, String currentCombination, List<String> combinations) {
if (currentCombination.length() == length) {
combinations.add(currentCombination);
return;
}
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (!currentCombination.contains(String.valueOf(c))) {
generateCombinationsHelper(input, length, currentCombination + c, combinations);
}
}
}
}2）C#实现：
csharp
using System;
using System.Collections.Generic;
namespace StringCombination
{
class Program
{
static void Main(string[] args)
{
string input = "中国浙江杭州";
int length = 6;
List<string> combinations = GenerateCombinations(input, length);
Console.WriteLine("满足条件的组合数量：" + combinations.Count);
Console.WriteLine("满足条件的组合：");
foreach (string combination in combinations)
{
Console.WriteLine(combination);
}
}
private static List<string> GenerateCombinations(string input, int length)
{
List<string> combinations = new List<string>();
GenerateCombinationsHelper(input, length, "", combinations);
return combinations;
}
private static void GenerateCombinationsHelper(string input, int length, string currentCombination, List<string> combinations)
{
if (currentCombination.Length == length)
{
combinations.Add(currentCombination);
return;
}
for (int i = 0; i < input.Length; i++)
{
char c = input[i];
if (!currentCombination.Contains(c.ToString()))
{
GenerateCombinationsHelper(input, length, currentCombination + c, combinations);
}
}
}
}
}3) python实现：
def generate_combinations(input_string, length):
combinations = []
generate_combinations_helper(input_string, length, "", combinations)
return combinations
def generate_combinations_helper(input_string, length, current_combination, combinations):
if len(current_combination) == length:
combinations.append(current_combination)
return
for i in range(len(input_string)):
char = input_string[i]
if char not in current_combination:
generate_combinations_helper(input_string, length, current_combination + char, combinations)
input_string = "上海金燕航空"
length = 6
combinations = generate_combinations(input_string, length)
print("满足条件的组合数量：", len(combinations))
print("满足条件的组合：")
for combination in combinations:
print(combination)4) object-c实现：
objective-c
#import <Foundation/Foundation.h>
void generateCombinationsHelper(NSString *input, int length, NSString *currentCombination, NSMutableArray *combinations) {
if (currentCombination.length == length) {
[combinations addObject:currentCombination];
return;
}
for (int i = 0; i < input.length; i++) {
unichar c = [input characterAtIndex:i];
NSString *charString = [NSString stringWithCharacters:&c length:1];
if (![currentCombination containsString:charString]) {
generateCombinationsHelper(input, length, [currentCombination stringByAppendingString:charString], combinations);
}
}
}
NSArray *generateCombinations(NSString *input, int length) {
NSMutableArray *combinations = [NSMutableArray array];
generateCombinationsHelper(input, length, @"", combinations);
return combinations;
}
int main(int argc, const char * argv[]) {
@autoreleasepool {
NSString *input = @"上海金燕航空";
int length = 6;
NSArray *combinations = generateCombinations(input, length);
NSLog(@"满足条件的组合数量：%lu", (unsigned long)combinations.count);
NSLog(@"满足条件的组合：");
for (NSString *combination in combinations) {
NSLog(@"%@", combination);
}
}
return 0;
}5) C++实现：
#include <iostream>
#include <vector>
using namespace std;
void generateCombinationsHelper(string input, int length, string currentCombination, vector<string>& combinations) {
if (currentCombination.length() == length) {
combinations.push_back(currentCombination);
return;
}
for (int i = 0; i < input.length(); i++) {
char c = input[i];
if (currentCombination.find(c) == string::npos) {
generateCombinationsHelper(input, length, currentCombination + c, combinations);
}
}
}
vector<string> generateCombinations(string input, int length) {
vector<string> combinations;
generateCombinationsHelper(input, length, "", combinations);
return combinations;
}
int main() {
string input = "上海金燕航空";
int length = 6;
vector<string> combinations = generateCombinations(input, length);
cout << "满足条件的组合数量：" << combinations.size() << endl;
cout << "满足条件的组合：" << endl;
for (string combination : combinations) {
cout << combination << endl;
}
return 0;
}6) c实现：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void generateCombinationsHelper(char* input, int length, char* currentCombination, char** combinations, int* count) {
if (strlen(currentCombination) == length) {
combinations[*count] = malloc((length + 1) * sizeof(char));
strcpy(combinations[*count], currentCombination);
(*count)++;
return;
}
for (int i = 0; i < strlen(input); i++) {
char c = input[i];
if (strchr(currentCombination, c) == NULL) {
char* newCombination = malloc((strlen(currentCombination) + 2) * sizeof(char));
strcpy(newCombination, currentCombination);
newCombination[strlen(newCombination)] = c;
newCombination[strlen(newCombination) + 1] = '\0';
generateCombinationsHelper(input, length, newCombination, combinations, count);
free(newCombination);
}
}
}
char** generateCombinations(char* input, int length, int* count) {
char** combinations = malloc(10000 * sizeof(char*));
*count = 0;
generateCombinationsHelper(input, length, "", combinations, count);
return combinations;
}
int main() {
char* input = "上海金燕航空";
int length = 6;
int count;
char** combinations = generateCombinations(input, length, &count);
printf("满足条件的组合数量：%d\n", count);
printf("满足条件的组合：\n");
for (int i = 0; i < count; i++) {
printf("%s\n", combinations[i]);
free(combinations[i]);
}
free(combinations);
return 0;
}