原文:
给定一个字符串，找到所有以括号形式断开给定字符串的方法。将每个子字符串括在括号内。
示例:
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)
我们强烈建议你尽量减少浏览器，先自己试试这个。 想法是用递归。我们维护两个参数——要处理的下一个字符的索引和到目前为止的输出字符串。我们从下一个要处理的字符的索引开始，将由未处理的字符串形成的子字符串追加到输出字符串中，并在剩余的字符串上递归，直到处理完整个字符串。我们使用 std::substr 来形成输出字符串。substr(pos，n)返回长度为 n 的子字符串，该子字符串从当前字符串的位置 pos 开始。
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(用绿色标记)和未处理的字符串(用红色标记)。
以下是上述想法的实现-
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 语言(一种计算机语言，尤用于创建网站)
// 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
计算机编程语言
# 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
output
(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(n2) t5】辅助空间: o(n 2
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