以括号形式打印所有断开字符串的方法

给定一个字符串，找到所有方法以括号形式打破给定字符串。将每个子字符串括在括号内。

例子：

Input : abc
Output: (a)(b)(c)
        (a)(bc)
        (ab)(c)
        (abc)


Input : abcd
Output : (a)(b)(c)(d)
         (a)(b)(cd)
         (a)(bc)(d)
         (a)(bcd)
         (ab)(c)(d)
         (ab)(cd)
         (abc)(d)
         (abcd)

我们强烈建议您最小化您的浏览器并首先自己尝试。
这个想法是使用递归。我们维护两个参数——下一个要处理的字符的索引和到目前为止的输出字符串。我们从要处理的下一个字符的索引开始，将未处理的字符串形成的子字符串附加到输出字符串并递归剩余的字符串，直到我们处理整个字符串。我们使用 std::substr 来形成输出字符串。 substr(pos, n) 返回从当前字符串的位置 pos 开始的长度为 n 的子字符串。

下图显示了输入字符串“abc”的递归树。图表上的每个节点显示已处理的字符串（以绿色标记）和未处理的字符串（以红色标记）。

下面是上述想法的实现——

C++

// C++ Program to find all combinations of Non-
// overlapping substrings formed from given
// string
#include 
using namespace std;
 
// find all combinations of non-overlapping
// substrings formed by input string str
// index – index of the next character to
//          be processed
// out - output string so far
void findCombinations(string str, int index, string out)
{
    if (index == str.length())
        cout << out << endl;
 
    for (int i = index; i < str.length(); i++)
    {
        // append substring formed by str[index,
        // i] to output string
        findCombinations(
            str,
            i + 1,
            out + "(" + str.substr(index, i + 1 - index)
                + ")");
    }
}
 
// Driver Code
int main()
{
    // input string
    string str = "abcd";
 
    findCombinations(str, 0, "");
 
    return 0;
}

Java

// Java program to find all combinations of Non-
// overlapping substrings formed from given
// string
 
class GFG
{
    // find all combinations of non-overlapping
    // substrings formed by input string str
    static void findCombinations(String str, int index,
                                 String out)
    {
        if (index == str.length())
            System.out.println(out);
  
        for (int i = index; i < str.length(); i++)
  
            // append substring formed by str[index,
            // i] to output string
            findCombinations(str, i + 1, out +
                "(" + str.substring(index, i+1) + ")" );
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        // input string
        String str = "abcd";
        findCombinations(str, 0, "");
    }
}
 
// Contributed by Pramod Kumar

Python3

# Python3 Program to find all combinations of Non-
# overlapping substrings formed from given
# string
 
# find all combinations of non-overlapping
# substrings formed by input string str
# index – index of the next character to
#          be processed
# out - output string so far
def findCombinations(string, index, out):
    if index == len(string):
        print(out)
 
    for i in range(index, len(string), 1):
 
        # append substring formed by str[index,
        # i] to output string
        findCombinations(string, i + 1, out + "(" +
                         string[index:i + 1] + ")")
 
# Driver Code
if __name__ == "__main__":
 
    # input string
    string = "abcd"
    findCombinations(string, 0, "")
 
# This code is contributed by
# sanjeev2552

C#

// C# program to find all combinations
// of Non-overlapping substrings formed
// from given string
using System;
 
class GFG {
    // find all combinations of non-overlapping
    // substrings formed by input string str
    public static void
    findCombinations(string str, int index, string @out)
    {
        if (index == str.Length) {
            Console.WriteLine(@out);
        }
 
        for (int i = index; i < str.Length; i++) {
 
            // append substring formed by
            // str[index, i] to output string
            findCombinations(
                str, i + 1,
                @out + "("
                    + str.Substring(index, (i + 1) - index)
                    + ")");
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        // input string
        string str = "abcd";
        findCombinations(str, 0, "");
    }
}
 
// This code is contributed by Shrikant13

输出

(a)(b)(c)(d)
(a)(b)(cd)
(a)(bc)(d)
(a)(bcd)
(ab)(c)(d)
(ab)(cd)
(abc)(d)
(abcd)

时间复杂度： O(N ² )
辅助空间： O(N ² )