打印字符串的所有子序列|迭代法

给定一个字符串s，打印出给定字符串的所有可能的子序列迭代的方式。我们已经讨论了递归方法来打印字符串的所有子序列。

例子：

Input : abc
Output : a, b, c, ab, ac, bc, abc

Input : aab
Output : a, b, aa, ab, aab

方法1：
在这里，我们讨论与Power Set相似的更容易，更简单的迭代方法。我们使用从1到2 ^ length – 1的二进制表示形式的位模式。

输入=“ abc”
考虑1到(2 ^ 3-1)，即1到7的二进制表示形式。
从二进制表示形式的左(MSB)到右侧(LSB)开始，并将与二进制表示形式的位值1相对应的输入字符串中的字符追加到Final子序列字符串sub中。

例子：
001 => abc。只有c对应于比特1。因此，子序列= c。
101 => abc。 a和c对应于比特1。因此，子序列= ac。
binary_representation(1)= 001 => c
binary_representation(2)= 010 => b
binary_representation(3)= 011 => bc
binary_representation(4)= 100 =>一个
binary_representation(5)= 101 =>交流电
binary_representation(6)= 110 => ab
binary_representation(7)= 111 => abc

下面是上述方法的实现：

C++

// C++ program to print all Subsequences
// of a string in iterative manner
#include 
using namespace std;
 
// function to find subsequence
string subsequence(string s, int binary, int len)
{
    string sub = "";
    for (int j = 0; j < len; j++)
 
        // check if jth bit in binary is 1
        if (binary & (1 << j))
 
            // if jth bit is 1, include it
            // in subsequence
            sub += s[j];
 
    return sub;
}
 
// function to print all subsequences
void possibleSubsequences(string s){
 
    // map to store subsequence
    // lexicographically by length
    map > sorted_subsequence;
 
    int len = s.size();
     
    // Total number of non-empty subsequence
    // in string is 2^len-1
    int limit = pow(2, len);
     
    // i=0, corresponds to empty subsequence
    for (int i = 1; i <= limit - 1; i++) {
         
        // subsequence for binary pattern i
        string sub = subsequence(s, i, len);
         
        // storing sub in map
        sorted_subsequence[sub.length()].insert(sub);
    }
 
    for (auto it : sorted_subsequence) {
         
        // it.first is length of Subsequence
        // it.second is set
        cout << "Subsequences of length = "
             << it.first << " are:" << endl;
              
        for (auto ii : it.second)
             
            // ii is iterator of type set
            cout << ii << " ";
         
        cout << endl;
    }
}
 
// driver function
int main()
{
    string s = "aabc";
    possibleSubsequences(s);
    return 0;
}

Java

// Java program to print all Subsequences
// of a String in iterative manner
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.SortedMap;
import java.util.TreeMap;
 
class Graph{
 
// Function to find subsequence
static String subsequence(String s,
                          int binary,
                          int len)
{
    String sub = "";
     
    for(int j = 0; j < len; j++)
     
        // Check if jth bit in binary is 1
        if ((binary & (1 << j)) != 0)
 
            // If jth bit is 1, include it
            // in subsequence
            sub += s.charAt(j);
 
    return sub;
}
 
// Function to print all subsequences
static void possibleSubsequences(String s)
{
     
    // Map to store subsequence
    // lexicographically by length
    SortedMap> sorted_subsequence = new TreeMap>();
 
    int len = s.length();
 
    // Total number of non-empty subsequence
    // in String is 2^len-1
    int limit = (int) Math.pow(2, len);
 
    // i=0, corresponds to empty subsequence
    for(int i = 1; i <= limit - 1; i++)
    {
         
        // Subsequence for binary pattern i
        String sub = subsequence(s, i, len);
         
        // Storing sub in map
        if (!sorted_subsequence.containsKey(sub.length()))
            sorted_subsequence.put(
                sub.length(), new HashSet<>());
            sorted_subsequence.get(
                sub.length()).add(sub);
    }
 
    for(Map.Entry> it : sorted_subsequence.entrySet())
    {
         
        // it.first is length of Subsequence
        // it.second is set
        System.out.println("Subsequences of length = " +
                           it.getKey() + " are:");
                            
        for(String ii : it.getValue())
         
            // ii is iterator of type set
            System.out.print(ii + " ");
 
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
    String s = "aabc";
     
    possibleSubsequences(s);
}
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 program to print all Subsequences
# of a string in iterative manner
 
# function to find subsequence
def subsequence(s, binary, length):
    sub = ""
    for j in range(length):
         
        # check if jth bit in binary is 1
        if (binary & (1 << j)):
 
            # if jth bit is 1, include it
            # in subsequence
            sub += s[j]
    return sub
 
# function to print all subsequences
def possibleSubsequences(s):
 
    # map to store subsequence
    # lexicographically by length
    sorted_subsequence = {}
 
    length = len(s)
     
    # Total number of non-empty subsequence
    # in string is 2^len-1
    limit = 2 ** length
     
    # i=0, corresponds to empty subsequence
    for i in range(1, limit):
         
        # subsequence for binary pattern i
        sub = subsequence(s, i, length)
         
        # storing sub in map
        if len(sub) in sorted_subsequence.keys():
            sorted_subsequence[len(sub)] = \
             tuple(list(sorted_subsequence[len(sub)]) + [sub])
        else:
            sorted_subsequence[len(sub)] = [sub]
 
    for it in sorted_subsequence:
         
        # it.first is length of Subsequence
        # it.second is set
        print("Subsequences of length =", it, "are:")
        for ii in sorted(set(sorted_subsequence[it])):
             
            # ii is iterator of type set
            print(ii, end = ' ')
        print()
 
# Driver Code
s = "aabc"
possibleSubsequences(s)
 
# This code is contributed by ankush_953

输出：

Subsequences of length = 1 are:
a b c 
Subsequences of length = 2 are:
aa ab ac bc 
Subsequences of length = 3 are:
aab aac abc 
Subsequences of length = 4 are:
aabc

时间复杂度： $O(2^{n} * l)$ ，其中n是查找子序列的字符串的长度，l是二进制字符串的长度。

方法2：
方法是获取最右边的设置位的位置，并在将给定字符串中的相应字符附加到子序列后重置该位，并将重复相同的操作，直到相应的二进制模式没有设置位为止。

如果输入为s =“ abc”
考虑1到(2 ^ 3-1)，即1到7的二进制表示形式。
001 => abc。只有c对应于比特1。因此，子序列= c
101 => abc。 a和c对应于比特1。因此，子序列= ac。
让我们使用5的二进制表示形式，即101。
最右边的位在位置1，在sub = c的开头附加字符，重置位置1 => 100
最右边的位在位置3，在sub = ac的开头附加字符，重置位置3 => 000
由于现在我们没有剩余的设置位，因此我们将停止计算子序列。

例子 ：
binary_representation(1)= 001 => c
binary_representation(2)= 010 => b
binary_representation(3)= 011 => bc
binary_representation(4)= 100 =>一个
binary_representation(5)= 101 =>交流电
binary_representation(6)= 110 => ab
binary_representation(7)= 111 => abc

下面是上述方法的实现：

C++

// C++ code all Subsequences of a
// string in iterative manner
#include 
using namespace std;
 
// function to find subsequence
string subsequence(string s, int binary)
{
    string sub = "";
    int pos;
     
    // loop while binary is greater than 0
    while(binary>0)
    {
        // get the position of rightmost set bit
        pos=log2(binary&-binary)+1;
         
        // append at beginning as we are
        // going from LSB to MSB
        sub=s[pos-1]+sub;
         
        // resets bit at pos in binary
        binary= (binary & ~(1 << (pos-1)));
    }
    reverse(sub.begin(),sub.end());
    return sub;
}
 
// function to print all subsequences
void possibleSubsequences(string s){
 
    // map to store subsequence
    // lexicographically by length
    map > sorted_subsequence;
 
    int len = s.size();
     
    // Total number of non-empty subsequence
    // in string is 2^len-1
    int limit = pow(2, len);
     
    // i=0, corresponds to empty subsequence
    for (int i = 1; i <= limit - 1; i++) {
         
        // subsequence for binary pattern i
        string sub = subsequence(s, i);
         
        // storing sub in map
        sorted_subsequence[sub.length()].insert(sub);
    }
 
    for (auto it : sorted_subsequence) {
         
        // it.first is length of Subsequence
        // it.second is set
        cout << "Subsequences of length = "
            << it.first << " are:" << endl;
             
        for (auto ii : it.second)
             
            // ii is iterator of type set
            cout << ii << " ";
         
        cout << endl;
    }
}
 
// driver function
int main()
{
    string s = "aabc";
    possibleSubsequences(s);
     
    return 0;
}

Python3

# Python3 program to print all Subsequences
# of a string in an iterative manner
from math import log2, floor
 
# function to find subsequence
def subsequence(s, binary):
    sub = ""
     
    # loop while binary is greater than
    while(binary > 0):
         
        # get the position of rightmost set bit
        pos=floor(log2(binary&-binary) + 1)
         
        # append at beginning as we are
        # going from LSB to MSB
        sub = s[pos - 1] + sub
         
        # resets bit at pos in binary
        binary= (binary & ~(1 << (pos - 1)))
 
    sub = sub[::-1]
    return sub
 
# function to print all subsequences
def possibleSubsequences(s):
 
    # map to store subsequence
    # lexicographically by length
    sorted_subsequence = {}
 
    length = len(s)
     
    # Total number of non-empty subsequence
    # in string is 2^len-1
    limit = 2 ** length
     
    # i=0, corresponds to empty subsequence
    for i in range(1, limit):
         
        # subsequence for binary pattern i
        sub = subsequence(s, i)
         
        # storing sub in map
        if len(sub) in sorted_subsequence.keys():
            sorted_subsequence[len(sub)] = \
            tuple(list(sorted_subsequence[len(sub)]) + [sub])
        else:
            sorted_subsequence[len(sub)] = [sub]
 
    for it in sorted_subsequence:
         
        # it.first is length of Subsequence
        # it.second is set
        print("Subsequences of length =", it, "are:")
        for ii in sorted(set(sorted_subsequence[it])):
             
            # ii is iterator of type set
            print(ii, end = ' ')
        print()
 
# Driver Code
s = "aabc"
possibleSubsequences(s)
 
# This code is contributed by ankush_953

输出：

Subsequences of length = 1 are:
a b c 
Subsequences of length = 2 are:
aa ab ac bc 
Subsequences of length = 3 are:
aab aac abc 
Subsequences of length = 4 are:
aabc

时间复杂度： $O(2^{n} * b)$ ，其中n是查找子序列的字符串的长度，b是二进制字符串中设置的位数。

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