如何打印表达式的所有可能的平衡括号？

Question

例如，对于元素 a,b,c,d，有 5 种可能的方法来获取相邻元素并将它们缩减为单个元素，其中必须一次组合两个元素（以下用括号表示）：

(((ab)c)d), ((a(bc))d), ((ab)(cd)), (a((bc)d)) and (a(b(cd)))

第一个示例乘以 a*b，然后乘以 c，然后乘以 d。第二个例子首先乘以b*c，然后乘以a，然后乘以d。

任何包含 2n 个元素的有效括号表达式都必须具有 n ( 和 n ) 属性，从左到右阅读，( 总是至少与 ) 一样多。

我知道对于 n 个数字，路数是第 (n-1) 个Catalan number。但是如何准确地生成所有结果分组呢？

谢谢

（顺便说一句：这个问题有超过 160 种等效公式，全部基于加泰罗尼亚数字枚举的不同组合对象。有关这些的最新列表，请参阅Richard Stanley's Catalan Addendum。）

Answer 1

这是 Python 中的实际代码，使用生成器来避免使用过多的内存。

#! /usr/bin/python

def parenthesized (exprs):
    if len(exprs) == 1:
        yield exprs[0]
    else:
        first_exprs = []
        last_exprs = list(exprs)
        while 1 < len(last_exprs):
            first_exprs.append(last_exprs.pop(0))
            for x in parenthesized(first_exprs):
                if 1 < len(first_exprs):
                    x = '(%s)' % x
                for y in parenthesized(last_exprs):
                    if 1 < len(last_exprs):
                        y = '(%s)' % y
                    yield '%s%s' % (x, y)

for x in parenthesized(['a', 'b', 'c', 'd']):
    print x

Answer 2

实际上4个元素的括号不止5个；您实际上并不是指“括号”。你真正要问的是 N 个元素可以有多少种不同的方式reduced，或者你可以用 N 个元素制作多少棵树，同时仍然保持它们的顺序。

这对应于将表达式精确细分 N-1 次。例如，在维基百科http://en.wikipedia.org/wiki/Catalan_number 文章中的这张图片中，如果我们有 4 个元素，则正好有 5 种方法可以对其应用二元运算符（需要正好有 3 个应用程序，因此正好有 3 个节点）：

例如，((a*b)*c)*d, (a*(b*c))*d, (a*b)*(c*d), a*((b*c)*d), a*(b*(c*d))

这里有一些简洁的python代码：

def associations(seq, **kw):
    """
        >>> associations([1,2,3,4])
        [(1, (2, (3, 4))), (1, ((2, 3), 4)), ((1, 2), (3, 4)), ((1, (2, 3)), 4), (((1, 2), 3), 4)] 
    """
    grouper = kw.get('grouper', lambda a,b:(a,b))
    lifter = kw.get('lifter', lambda x:x)

    if len(seq)==1:
        yield lifter(seq[0])
    else:
        for i in range(len(seq)):
            left,right = seq[:i],seq[i:] # split sequence on index i

            # return cartesian product of left x right
            for l in associations(left,**kw):
                for r in associations(right,**kw):
                    yield grouper(l,r)

请注意如何用此代码替换有趣的函数 grouper，例如grouper=list，或grouper=Tree，或grouper=...。

演示：

for assoc in associations('abcd'):
    print assoc

('a', ('b', ('c', 'd')))
('a', (('b', 'c'), 'd'))
(('a', 'b'), ('c', 'd'))
(('a', ('b', 'c')), 'd')
((('a', 'b'), 'c'), 'd')

Answer 3

使用递归

   for each balanced expression of n-1 parentheses 
     for each pos i from 0 to m of an expression
       add '('
       for each pos  j from i + 1 to m
          add ')' if expression is balanced

您将得到的顺序如下：

n=0: 

n=1: ()

n=2: []() , [()]

n=3: {}[]() , {[]}() , {[]()} , {}[()] , {[()]}

这里我每次都会更改括号 (,[,{ 以突出显示算法的工作原理。

Answer 4

递归将是要走的路

将abcd拆分成

(a) (bcd)
(ab) (cd)
(abc) (d)

这些是一些可能性

现在您可以递归地拆分每个字符串（拆分时忽略括号）说（bcd）一种可能性

(b) (cd)

所以现在另一种组合是

((a)(b)(cd))

一旦你得到一个只有一个字母的字符串，你就可以停止递归树

Answer 5

import java.util.ArrayList;


public class Parantheses {

    private ArrayList<String> parStringList;
    private int total;

    public void exploreParantheses(String parString,int leftCnt,int rightCnt)
    {
        if (leftCnt < total)
        {
            exploreParantheses(parString + "(", leftCnt+1, rightCnt);
        }
        if ((rightCnt < total) &&(rightCnt < leftCnt))
        {
            exploreParantheses(parString + ")", leftCnt, rightCnt+1);
        }
        else if (rightCnt == total)
        {
            parStringList.add(parString);
        }
    }
    public ArrayList<String> generateParanthesis(int numPars)
    {
        this.total = numPars;
        this.parStringList = new ArrayList<String>();
        exploreParantheses("", 0, 0);
        return this.parStringList;
    }
    public static void main(String args[])
    {
        ArrayList<String> par;
        par = (new Parantheses()).generateParanthesis(6);
        for (String str: par)
            System.out.println(str);
    }
}

Answer 6

*

**Run this to generate all balanced parantheses:
//sudosuhan
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#define MAX_SIZE 200
void _printParenthesis(int pos, int n1, int open1, int close1, int n2, int open2, int close2, int n3, int open3, int close3);
void printParenthesis(int n1 , int n2 , int n3 )
{
  if(n1 > 0 || n2 > 0 || n3 > 0)
     _printParenthesis(0, n1, 0, 0, n2, 0, 0, n3, 0, 0);
  return;
}
void _printParenthesis(int pos, int n1, int open1, int close1, int n2, int open2, int close2, int n3, int open3, int close3)
{
  static char str[MAX_SIZE];     

  if(close1 == n1 && close2 == n2 && close3 == n3 ) 
  {
    printf("%s \n", str);
    return;
  }
  else
  {
    bool run1 = open1 > close1;
    bool run2 = open2 > close2;
    bool run3 = open3 > close3;
    if(run3)
    {
      str[pos] = ')';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2, n3, open3, close3+1);

      if(open3 < n3)
      {
      str[pos] = '(';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2, n3, open3+1, close3);
      }
    }
    else if(run2 && !run3)
    {
      str[pos] = '}';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2+1, n3, open3, close3);

      if(open3 < n3)
      {
      str[pos] = '(';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2, n3, open3+1, close3);
      }
      if(open2 < n2)
      {
      str[pos] = '{';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2+1, close2, n3, open3, close3);
      }
    }
    else if(run1 && !run2 && !run3)
    {
      str[pos] = ']';
      _printParenthesis(pos+1, n1, open1, close1+1, n2, open2, close2, n3, open3, close3);

      if(open3 < n3)
      {
      str[pos] = '(';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2, n3, open3+1, close3);
      }
      if(open2 < n2)
      {
      str[pos] = '{';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2+1, close2, n3, open3, close3);
      }
      if(open1 < n1)
      {
      str[pos] = '[';
      _printParenthesis(pos+1, n1, open1+1, close1, n2, open2, close2, n3, open3, close3);
      }
    }
    else if(!run1 && !run2 && !run3)
    {
      if(open3 < n3)
      {
       str[pos] = '(';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2, close2, n3, open3+1, close3);
      }
      if(open2 < n2)
      {
      str[pos] = '{';
      _printParenthesis(pos+1, n1, open1, close1, n2, open2+1, close2, n3, open3, close3);
      }
      if(open1 < n1)
      {
      str[pos] = '[';
      _printParenthesis(pos+1, n1, open1+1, close1, n2, open2, close2, n3, open3, close3);
      }
    }

  }
}
/* driver program to test above functions */
int main()
{
  int n1, n2, n3;
  n1 = 6;
  n2 = 1;
  n3 = 1;
  printParenthesis(n1, n2, n3);
  return 0;
}**

*

Answer 7

还有，这里有一些相同的 C++ 代码：

bool is_a_solution( string partial,int n,int k) { 
if(partial.length() == n*2 )  
 return true;
return false;
  }
string constructCandidate(int n,string input,string partial, int k) { 
int xcount=0,ycount=0;
int count;
int i;
string candi;
if(k == 0)  
    return string("(");
else { 
    for(i=0;i<partial.length();i++)  { 
        if( partial[i] == '(') xcount++;
        if( partial[i] == ')') ycount++;
    }
    if( xcount  <n) candi+="(";  
    if( ycount < xcount) candi+=")"; 
    }
return candi;}                                                                                void backTrack(int n,string input, string partial,int k )  { 
int i, numCanditate;
string mypartial;
    if( is_a_solution(partial,n,k)) { 
    cout <<partial<<"\n"; 
 }else {
     string  candi=constructCandidate(n,input,partial,k); 
     for(i=0;i<candi.length();i++) {
         backTrack(n,input,partial+candi[i],k+1); 
     }
 }
void paranthesisPrint(int n){
   backTrack(n,"()", "",0); 
   }

Answer 8

这是从给定的 n+1 因子生成所有可能的平衡括号字符串的 C# 版本。

注意问题的解决方案基本满足加泰罗尼亚递归关系（详细可参考http://codingworkout.blogspot.com/2014/08/all-possible-paranthesis.html,http://en.wikipedia.org/wiki/Catalan_number）

string[] CatalanNumber_GeneratingParanthesizedFactorsRecursive(string s)
        {
            if(s.Length == 1)
            {
                return new string[] {s};
            }
            if(s.Length == 2)
            {
                string r = "(" + s + ")";
                return new string[] { r };
            }
            List<string> results = new List<string>();
            for (int i = 1; i < s.Length; i++)
            {
                var r1 = this.CatalanNumber_GeneratingParanthesizedFactorsRecursive(
                    s.Substring(0, i));
                var r2 = this.CatalanNumber_GeneratingParanthesizedFactorsRecursive(
                    s.Substring(i));
                foreach(var s1 in r1)
                {
                    foreach(var s2 in r2)
                    {
                        string r = "(" + s1 + s2 + ")";
                        results.Add(r);
                    }
                }
            }
            return results.ToArray();
        }

在哪里

string[] CatalanNumber_GeneratingParanthesizedFactors(string s)
        {
            s.ThrowIfNullOrWhiteSpace("s");
            if(s.Length == 1)
            {
                return new string[] {s};
            }
            var r = this.CatalanNumber_GeneratingParanthesizedFactorsRecursive(
                s);
            return r;
        }

单元测试是

[TestMethod]
        public void CatalanNumber_GeneratingParanthesizedFactorsTests()
        {
            var CatalanNumbers = new int[] { 1, 1, 2, 5, 14, 42, 132, 429, 
                1430, 4862, 16796 };
            string input = "";
            for (int i = 1; i <= 10; i++)
            {
                input += i;
                var results2 = this.CatalanNumber_GeneratingParanthesizedFactors(input);
                Assert.AreEqual(results2.Length, CatalanNumbers[input.Length-1]);
                Debug.WriteLine("-----------------------------------------------");
                foreach (string ss in results2)
                {
                    Debug.WriteLine(ss);
                }
            }
            string s = "a";
            var r = this.CatalanNumber_GeneratingParanthesizedFactors(s);
            Assert.AreEqual(r.Length, 1);
            Assert.AreEqual(s, r[0]);
            s = "ab";
            r = this.CatalanNumber_GeneratingParanthesizedFactors(s);
            Assert.AreEqual("(ab)", r[0]);
            s = "abc";
            r = this.CatalanNumber_GeneratingParanthesizedFactors(s);
            string[] output = new string[] { "(a(bc))", "((ab)c)" };
            Assert.AreEqual(2, r.Length);
            foreach(var o in output)
            {
                Assert.AreEqual(1, r.Where(rs => (rs == o)).Count());
            }
            s = "abcd";
            r = this.CatalanNumber_GeneratingParanthesizedFactors(s);

            output = new string[] { "(a(b(cd)))", "(a((bc)d))", "((ab)(cd))", "(((ab)c)d)", "((a(bc))d)"};
            Assert.AreEqual(5, r.Length);
            foreach (var o in output)
            {
                Assert.AreEqual(1, r.Where(rs => (rs == o)).Count());
            }
        }

Answer 9

最初的左括号有一个唯一匹配的右括号，这样两个括号之间的和后面的都是有效的表达式。这导致了这里用 Scala 表示的简单递归解决方案。

def catalan(n: Int): List[String] =
    if (n == 0) List("")
    else
        for {
            k <- (0 to n - 1).toList
            first <- catalan(k)
            rest <- catalan(n - 1 - k)
        } yield "a" + first + "b" + rest

这里我用“a”表示左括号，“b”表示右括号。

catalan(0)               List()
catalan(1)               List(ab)
catalan(2)               List(abab, aabb)
catalan(3)               List(ababab, abaabb, aabbab, aababb, aaabbb) 
catalan(5).size          42