找到格式正确的括号的所有组合

时间:2009-04-07 21:43:20

标签: c# algorithm f# catalan

这是在与朋友交谈时提出的,我想我会问这里,因为这是一个有趣的问题,并希望看到其他人的解决方案。

任务是编写一个函数Brackets(int n),它打印1 ... n中格式良好括号的所有组合。对于Brackets(3),输出将是

()
(())  ()()   
((()))  (()())  (())()  ()(())  ()()()

27 个答案:

答案 0 :(得分:49)

对它采取了一个裂缝.. C#也。

public void Brackets(int n) {
    for (int i = 1; i <= n; i++) {
        Brackets("", 0, 0, i);
    }
}

private void Brackets(string output, int open, int close, int pairs) {
    if((open==pairs)&&(close==pairs)) {
        Console.WriteLine(output);
    } else {
        if(open<pairs)
            Brackets(output + "(", open+1, close, pairs);
        if(close<open)
            Brackets(output + ")", open, close+1, pairs);
    }
}

递归正在利用这样一个事实:你永远不能添加比所需数量的对更多的开括号,并且你永远不能添加更多的结束括号而不是开括号..

答案 1 :(得分:9)

F#

这是一个解决方案,与我之前的解决方案不同,我认为可能是正确的。此外,它更有效。

#light

let brackets2 n =
    let result = new System.Collections.Generic.List<_>()
    let a = Array.create (n*2) '_'
    let rec helper l r diff i =
        if l=0 && r=0 then
            result.Add(new string(a))
        else
            if l > 0 then
                a.[i] <- '('
                helper (l-1) r (diff+1) (i+1)
            if diff > 0 then
                a.[i] <- ')'
                helper l (r-1) (diff-1) (i+1)
    helper n n 0 0
    result

示例:

(brackets2 4) |> Seq.iter (printfn "%s")

(*
(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())(())
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()
*)

答案 2 :(得分:8)

可能的组合数是N对C(n)的Catalan number

这个问题discussed on the joelonsoftware.com forums非常复杂,包括迭代,递归和迭代/位移解决方案。那里有一些很酷的东西。

以下是C#论坛中建议的快速递归解决方案:

C#

public void Brackets(int pairs) {
    if (pairs > 1) Brackets(pairs - 1);
    char[] output = new char[2 * pairs];

    output[0] = '(';
    output[1] = ')';

    foo(output, 1, pairs - 1, pairs, pairs);
    Console.writeLine();
}

public void foo(char[] output, int index, int open, int close,
        int pairs) {
    int i;

    if (index == 2 * pairs) {
        for (i = 0; i < 2 * pairs; i++)
            Console.write(output[i]);
        Console.write('\n');
        return;
    }

    if (open != 0) {
        output[index] = '(';
        foo(output, index + 1, open - 1, close, pairs);
    }

    if ((close != 0) && (pairs - close + 1 <= pairs - open)) {
        output[index] = ')';
        foo(output, index + 1, open, close - 1, pairs);
    }

    return;
}

支架(3);

  

输出:
  ()
  (())()()
  ((()))(()())(())()()(())()()()

答案 3 :(得分:8)

第一个投票答案的Python版本。

def foo(output, open, close, pairs):
    if open == pairs and close == pairs:
        print output
    else:
        if open<pairs:
            foo(output+'(', open+1, close, pairs)
        if close<open:
            foo(output+')', open, close+1, pairs)

foo('', 0, 0, 3)

答案 4 :(得分:5)

这是另一种F#解决方案,偏好优雅而非效率,尽管备忘可能会导致性能相对较好的变体。

let rec parens = function
| 0 -> [""]
| n -> [for k in 0 .. n-1 do
        for p1 in parens k do
        for p2 in parens (n-k-1) ->
          sprintf "(%s)%s" p1 p2]

同样,这只会产生一个包含完全 n对parens(而不是最多n)的字符串的列表,但它很容易包装。

答案 5 :(得分:4)

C ++中的简单解决方案:

#include <iostream>
#include <string>

void brackets(string output, int open, int close, int pairs)
{
    if(open == pairs && close == pairs)
            cout << output << endl;
    else
    {
            if(open<pairs)
                    brackets(output+"(",open+1,close,pairs);
            if(close<open)
                    brackets(output+")",open,close+1,pairs);
    }
}

int main()
{
    for(int i=1;i<=3;i++)
    {
            cout << "Combination for i = " << i << endl;
            brackets("",0,0,i);
    }
}

<强>输出:

Combination for i = 1
()
Combination for i = 2
(())
()()
Combination for i = 3
((()))
(()())
(())()
()(())
()()()

答案 6 :(得分:4)

F#

更新:这个答案错了​​。我的N = 4未命中,例如“(())(())”。 (你知道为什么吗?)我将很快发布一个正确的(更有效的)算法。

(对你们所有的选民感到羞耻,因为没有抓住我!:)


效率低,但简短而简单。 (请注意,它只打印'nth'行;从1..n循环调用以获取问题所要求的输出。)

#light
let rec brackets n =
    if n = 1 then
        ["()"]
    else
        [for s in brackets (n-1) do
            yield "()" ^ s
            yield "(" ^ s ^ ")"
            yield s ^ "()"]

示例:

Set.of_list (brackets 4) |> Set.iter (printfn "%s")
(*
(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()
*)

答案 7 :(得分:3)

该死的 - 每个人都打败了我,但我有一个很好的工作示例:)

http://www.fiveminuteargument.com/so-727707

关键是确定规则,实际上非常简单:

  • 构建字符串char-by-char
  • 在字符串中的给定点
    • 如果到目前为止字符串中的括号(包括空str),请添加一个空心括号并递归
    • 如果使用了所有打开的括号,请添加一个小括号并递归
    • 否则,递归两次,每种类型的支架一次
  • 当你走到尽头时停止: - )

答案 8 :(得分:3)

Common Lisp:

这不会打印它们,但会生成所有可能结构的列表列表。我的方法与其他方法有点不同。它将brackets(n - 1)的解决方案重新构建为brackets(n)。我的解决方案不是尾递归,但可以通过一些工作来实现。

代码

(defun brackets (n)
  (if (= 1 n)
      '((()))
      (loop for el in (brackets (1- n))
            when (cdr el)
            collect (cons (list (car el)) (cdr el))
            collect (list el)
            collect (cons '() el))))

答案 9 :(得分:2)

简单的F#/ OCaml解决方案:

let total_bracket n =
    let rec aux acc = function
        | 0, 0 -> print_string (acc ^ "\n")
        | 0, n -> aux (acc ^ ")") (0, n-1)
        | n, 0 -> aux (acc ^ "(") (n-1, 1)
        | n, c ->
                aux (acc ^ "(") (n-1, c+1);
                aux (acc ^ ")") (n,   c-1)
    in
    aux "" (n, 0)

答案 10 :(得分:1)

为什么不能这么简单,这个想法很简单

括号(n) - &gt; '()'+括号(n-1)0                 '('+括号(n-1)+')'0                 括号(n-1)+'()'

其中0是上面的连接操作

public static Set<String> brackets(int n) {
    if(n == 1){
        Set<String> s = new HashSet<String>();
        s.add("()");
        return s;
    }else{
        Set<String> s1 = new HashSet<String>();
        Set<String> s2 = brackets(n - 1);
        for(Iterator<String> it = s2.iterator(); it.hasNext();){
            String s = it.next();
            s1.add("()" + s);
            s1.add("(" + s + ")");
            s1.add(s + "()");
        }
        s2.clear();
        s2 = null;
        return s1;
    }
}

答案 11 :(得分:1)

基于递归回溯算法的提供商C#版本,希望它有所帮助。

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {

    UIBarButtonItem.appearance().setTitleTextAttributes([NSAttributedStringKey.font: UIFont(name: "SF Pro Display", size: 17)!], for: .normal)

    return true
}

答案 12 :(得分:1)

Groovy版本基于上面的markt优雅的c#解决方案。动态检查打开和关闭(信息在输出和args中重复)以及删除一些无关的逻辑检查。

3.times{ 
    println bracks(it + 1)
}

def bracks(pairs, output=""){
    def open = output.count('(')
    def close = output.count(')')

    if (close == pairs) {
        print "$output "
    }
    else {
        if (open < pairs) bracks(pairs, "$output(")
        if (close < open) bracks(pairs, "$output)")
    }
    ""
}

答案 13 :(得分:1)

Haskell中:

我试图想出一个优雅的monad-y方式:

import Control.Applicative

brackets :: Int -> [String]
brackets n = f 0 0 where
    f pos depth =
        if pos < 2*n
            then open <|> close
            else stop where
                -- Add an open bracket if we can
                open =
                    if depth < 2*n - pos
                        then ('(' :) <$> f (pos+1) (depth+1)
                        else empty

                -- Add a closing bracket if we can
                close = 
                    if depth > 0
                        then (')' :) <$> f (pos+1) (depth-1)
                        else empty

                -- Stop adding text.  We have 2*n characters now.
                stop = pure ""

main = readLn >>= putStr . unlines . brackets

答案 14 :(得分:1)

def @memo brackets ( n )
    => [] if n == 0 else around( n ) ++ pre( n ) ++ post( n ) ++ [ "()" * n) ]

def @memo pre ( n )
    => map ( ( s ) => "()" ++ s, pre ( n - 1 ) ++ around ( n - 1 ) ) if n > 2 else []

def @memo post ( n )
    => map ( ( s ) => s ++ "()", post ( n - 1 ) ++ around ( n - 1 ) ) if n > 2 else []

def @memo around ( n )
    => map ( ( s ) => "(" ++ s ++ ")", brackets( n - 1 ) )

kin,这类似于基于演员模型的线性python与特征。我没有完成实现@memo,但上述工作没有优化)

答案 15 :(得分:1)

这是C ++中的解决方案。我使用的主要思想是我从前面的 i (其中 i 是括号对的数量)的输出,并将其作为输入提供给下一个< EM> I 的。然后,对于输入中的每个字符串,我们在字符串中的每个位置放置一个括号对。新的字符串被添加到集合中以消除重复。

#include <iostream>
#include <set>
using namespace std;
void brackets( int n );
void brackets_aux( int x, const set<string>& input_set, set<string>& output_set );

int main() {
    int n;
    cout << "Enter n: ";
    cin >> n;
    brackets(n);
    return 0;
}

void brackets( int n ) {
    set<string>* set1 = new set<string>;
    set<string>* set2;

    for( int i = 1; i <= n; i++ ) {
        set2 = new set<string>;
        brackets_aux( i, *set1, *set2 );
        delete set1;
        set1 = set2;
    }
}

void brackets_aux( int x, const set<string>& input_set, set<string>& output_set ) {
    // Build set of bracket strings to print
    if( x == 1 ) {
        output_set.insert( "()" );
    }
    else {
        // For each input string, generate the output strings when inserting a bracket pair
        for( set<string>::iterator s = input_set.begin(); s != input_set.end(); s++ ) {
            // For each location in the string, insert bracket pair before location if valid
            for( unsigned int i = 0; i < s->size(); i++ ) {
                string s2 = *s;
                s2.insert( i, "()" );
                output_set.insert( s2 );
            }
            output_set.insert( *s + "()" );
        }
    }

    // Print them
    for( set<string>::iterator i = output_set.begin(); i != output_set.end(); i++ ) {
        cout << *i << "  ";
    }
    cout << endl;
}

答案 16 :(得分:0)

不是最优雅的解决方案,但这就是我在C ++(Visual Studio 2008)中的表现。利用STL集来消除重复,我只是天真地在上一代的每个字符串中的每个字符串索引中插入new()对,然后递归。

#include "stdafx.h"
#include <iostream>
#include <string>
#include <set>

using namespace System;
using namespace std;

typedef set<string> StrSet;

void ExpandSet( StrSet &Results, int Curr, int Max )
{
    if (Curr < Max)
    {
        StrSet NewResults;

        for (StrSet::iterator it = Results.begin(); it != Results.end(); ++it)
        {
            for (unsigned int stri=0; stri < (*it).length(); stri++)
            {
                string NewStr( *it );
                NewResults.insert( NewStr.insert( stri, string("()") ) );
            }
        }
        ExpandSet( NewResults, Curr+1, Max );

        Results = NewResults;
    }
}    

int main(array<System::String ^> ^args)
{
    int ParenCount = 0;

    cout << "Enter the parens to balance:" << endl;
    cin  >> ParenCount;

    StrSet Results;
    Results.insert( string("()") );

    ExpandSet(Results, 1, ParenCount);

    cout << Results.size() << ": Total # of results for " << ParenCount << " parens:" << endl;

    for (StrSet::iterator it = Results.begin(); it != Results.end(); ++it)
    {
        cout << *it << endl;
    }


    return 0;
}

答案 17 :(得分:0)

//C program to print all possible n pairs of balanced parentheses  


#include<stdio.h>

void fn(int p,int n,int o,int c);

void main()
{
    int n;
    printf("\nEnter n:");
    scanf("%d",&n);
    if(n>0)  
        fn(0,n,0,0);
}

void fn(int p,int n,into,int c)
{  
    static char str[100];
    if(c==n)
    {
        printf("%s\n",str);
        return;
    }
    else
    {
        if(o>c)
        {
            str[p]='}';
            fn(p+1,n,o,c+1);
        }
        if(o<n)
        {
            str[p]='{';
            fn(p+1,n;o+1,c);
        }
    }
}

答案 18 :(得分:0)

ruby​​版本:

def foo output, open, close, pairs
  if open == pairs and close == pairs
      p output
  else
    foo(output + '(', open+1, close, pairs) if open < pairs
    foo(output + ')', open, close+1, pairs) if close < open
  end
end
foo('', 0, 0, 3)

答案 19 :(得分:0)

public static void printAllValidBracePermutations(int size) {
    printAllValidBracePermutations_internal("", 0, 2 * size);
}

private static void printAllValidBracePermutations_internal(String str, int bal, int len) {
    if (len == 0) System.out.println(str);
    else if (len > 0) {
        if (bal <= len / 2) printAllValidBracePermutations_internal(str + "{", bal + 1, len - 1);
        if (bal > 0) printAllValidBracePermutations_internal(str + "}", bal - 1, len - 1);
    }
}

答案 20 :(得分:0)

另一个低效但优雅的答案=&gt;

public static Set<String> permuteParenthesis1(int num)
{   
    Set<String> result=new HashSet<String>();
    if(num==0)//base case
        {
            result.add("");
            return result;
        }
    else
        {
            Set<String> temp=permuteParenthesis1(num-1); // storing result from previous result.
            for(String str : temp)
            {
                for(int i=0;i<str.length();i++)
                {
                    if(str.charAt(i)=='(')
                    {
                        result.add(insertParen(str, i)); // addinng `()` after every left parenthesis.
                    }
                }
                result.add("()"+str); // adding "()" to the beginning.
            }

        }
    return result;


}
public static String insertParen(String str,int leftindex)
{
    String left=str.substring(0, leftindex+1);
    String right=str.substring(leftindex+1);
    return left+"()"+right;
}

public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println(permuteParenthesis1(3));

}

答案 21 :(得分:0)

尝试记忆:

void push_strings(int i, int j ,vector<vector <string>> &T){
    for (int k=0; k< T[j].size(); ++k){
        for (int l=0; l< T[i - 1 - j].size(); ++l){
            string s = "(" + T[j][k] + ")" + T[i-1 - j][l];
            T[i].push_back(s);
        }
    }
}

vector<string> generateParenthesis(int n) {
    vector<vector <string>> T(n+10);
    T[0] = {""};

    for (int i =1; i <=n; ++i){
        for(int j=0; j<i; ++j){
            push_strings(i,j, T);
        }
    }

    return T[n];
}

答案 22 :(得分:0)

def form_brackets(n: int) -> set:
    combinations = set()
    if n == 1:
        combinations.add('()')
    else:
        previous_sets = form_brackets(n - 1)
        for previous_set in previous_sets:
            for i, c in enumerate(previous_set):
                temp_string = "{}(){}".format(previous_set[:i+1], previous_set[i+1:])
                combinations.add(temp_string)

    return combinations

答案 23 :(得分:0)

void function(int n, string str, int open, int close)
{
    if(open>n/2 || close>open)
        return;
    if(open==close && open+close == n)
    {
        cout<<" "<<str<<endl;
        return;
    }
    function(n, str+"(", open+1, close);
    function(n, str+")", open, close+1);
}

来电者 - function(2*brackets, str, 0, 0);

答案 24 :(得分:0)

results = []
num = 0

def print_paratheses(left, right):
    global num
    global results

    # When nothing left, print the results.
    if left == 0 and right == 0:
        print results
        return

    # pos is the next postion we should insert parenthesis.
    pos = num - left - right
    if left > 0:
        results[pos] = '('
        print_paratheses(left - 1, right)

    if left < right:
        results[pos] = ')'
        print_paratheses(left, right - 1)

def print_all_permutations(n):
    global num
    global results
    num = n * 2
    results = [None] * num
    print_paratheses(n, n)

参考:Permutations of Parentheses

答案 25 :(得分:0)

在javascript / nodejs中。

该程序最初旨在回答“终极问题”,但是枚举有效的括号组合非常完美。

function* life(universe){
    if( !universe ) yield '';
    for( let everything = 1 ; everything <= universe ; ++everything )
        for( let meaning of life(everything - 1) )
            for( let question of life(universe - everything) )    
                yield question + '(' + meaning + ')';
}
let love = 5;
let answer = [...life(love)].length;
console.log(answer);

function brackets(n){
    for( k = 1 ; k <= n ; k++ ){
        console.log(...life(k));
    }
}

brackets(5);

答案 26 :(得分:-1)

在C#中

    public static void CombiParentheses(int open, int close, StringBuilder str)
    {
        if (open == 0 && close == 0)
        {
            Console.WriteLine(str.ToString());
        }
        if (open > 0) //when you open a new parentheses, then you have to close one parentheses to balance it out.
        {                
            CombiParentheses(open - 1, close + 1, str.Append("{"));
        }
        if (close > 0)
        {                
            CombiParentheses(open , close - 1, str.Append("}"));
        }
    }