给定长度为M的数字字符串S和整数N ，任务是找到最大为N的S的所有不同组合(允许重复)。
例子：
Input: S = “124”, N = 100
Output: 1, 11, 12, 14, 2, 21, 22, 24, 4, 41, 42, 44
Explanation: Combinations “111”, “112”, “122”, “124”, “412” are greater than 100. Therefore, these combinations are excluded from the output.
Input: S = “345”, N = 400
Output: 3, 33, 333, 334, 335, 34, 343, 344, 345, 35, 353, 354, 355, 4, 43, 44, 45, 5, 53, 54, 55
方法：想法是使用回溯生成所有可能的数字，然后打印不超过N的那些数字。请按照以下步骤解决问题：
- 初始化一组字符串(例如， combinations [])，以存储S的所有不同组合，这些组合在数字上不超过N。
- 初始化一个字符串ans，以存储可能来自S的数字的当前组合。
- 声明一个函数generateCombinations()以生成其值小于给定值N的所有必需组合，并且该函数定义为：
- 使用变量i遍历[0，M]范围内的字符串S并执行以下操作：
- 将当前字符S [i]推入ans并将当前字符串ans转换为数字并将其存储在x中。
- 如果x小于或等于N，则将字符串ans推入groups []并递归调用函数generateCombinations() 。
- 通过从ans中删除第i个字符，将其返回到先前的状态。
- 使用变量i遍历[0，M]范围内的字符串S并执行以下操作：
- 完成上述步骤后，打印所有以combinations []存储的字符串的集合。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Store the current sequence of s
string combination;
// Store the all the required sequences
set combinations;
// Function to print all sequences of S
// satisfying the required condition
void printSequences(
set combinations)
{
// Print all strings in the set
for (string s : combinations) {
cout << s << ' ';
}
}
// Function to generate all sequences
// of string S that are at most N
void generateCombinations(string& s, int n)
{
// Iterate over string s
for (int i = 0; i < s.size(); i++) {
// Push ith character to combination
combination.push_back(s[i]);
// Convert the string to number
long x = stol(combination);
// Check if the condition is true
if (x <= n) {
// Push the current string to
// the final set of sequences
combinations.insert(combination);
// Recursively call function
generateCombinations(s, n);
}
// Backtrack to its previous state
combination.pop_back();
}
}
// Driver Code
int main()
{
string S = "124";
int N = 100;
// Function Call
generateCombinations(S, N);
// Print required sequences
printSequences(combinations);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Store the current sequence of s
static String combination = "";
// Store the all the required sequences
static HashSet combinations = new LinkedHashSet();
// Function to print all sequences of S
// satisfying the required condition
static void printSequences(
HashSet combinations)
{
// Print all Strings in the set
for(String s : combinations)
{
System.out.print(s + " ");
}
}
// Function to generate all sequences
// of String S that are at most N
static void generateCombinations(String s, int n)
{
// Iterate over String s
for(int i = 0; i < s.length(); i++)
{
// Push ith character to combination
combination += (s.charAt(i));
// Convert the String to number
long x = Integer.valueOf(combination);
// Check if the condition is true
if (x <= n)
{
// Push the current String to
// the final set of sequences
combinations.add(combination);
// Recursively call function
generateCombinations(s, n);
}
// Backtrack to its previous state
combination = combination.substring(
0, combination.length() - 1);
}
}
// Driver Code
public static void main(String[] args)
{
String S = "124";
int N = 100;
// Function Call
generateCombinations(S, N);
// Print required sequences
printSequences(combinations);
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program for the above approach
# Store the current sequence of s
combination = "";
# Store the all the required sequences
combinations = [];
# Function to print all sequences of S
# satisfying the required condition
def printSequences(combinations) :
# Print all strings in the set
for s in (combinations) :
print(s, end = ' ');
# Function to generate all sequences
# of string S that are at most N
def generateCombinations(s, n) :
global combination;
# Iterate over string s
for i in range(len(s)) :
# Push ith character to combination
combination += s[i];
# Convert the string to number
x = int(combination);
# Check if the condition is true
if (x <= n) :
# Push the current string to
# the final set of sequences
combinations.append(combination);
# Recursively call function
generateCombinations(s, n);
# Backtrack to its previous state
combination = combination[:-1];
# Driver Code
if __name__ == "__main__" :
S = "124";
N = 100;
# Function Call
generateCombinations(S, N);
# Print required sequences
printSequences(combinations);
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Store the current sequence of s
static String combination = "";
// Store the all the required sequences
static SortedSet combinations = new SortedSet();
// Function to print all sequences of S
// satisfying the required condition
static void printSequences(
SortedSet combinations)
{
// Print all Strings in the set
foreach(String s in combinations)
{
Console.Write(s + " ");
}
}
// Function to generate all sequences
// of String S that are at most N
static void generateCombinations(String s, int n)
{
// Iterate over String s
for(int i = 0; i < s.Length; i++)
{
// Push ith character to combination
combination += (s[i]);
// Convert the String to number
long x = Int32.Parse(combination);
// Check if the condition is true
if (x <= n)
{
// Push the current String to
// the readonly set of sequences
combinations.Add(combination);
// Recursively call function
generateCombinations(s, n);
}
// Backtrack to its previous state
combination = combination.Substring(
0, combination.Length - 1);
}
}
// Driver Code
public static void Main(String[] args)
{
String S = "124";
int N = 100;
// Function Call
generateCombinations(S, N);
// Print required sequences
printSequences(combinations);
}
}
// This code is contributed by 29AjayKumar 输出：
1 11 12 14 2 21 22 24 4 41 42 44时间复杂度： O(N N )
辅助空间： O(N N )