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
()
(()) ()()
((())) (()()) (())() ()(()) ()()()
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:
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:强>
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:
答案 8 :(得分:3)
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。
(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, "$output)")
}
""
}
答案 13 :(得分:1)
我试图想出一个优雅的monad-y方式:
import Control.Applicative
brackets :: Int -> [String]
brackets n = f 0 0 where
f pos depth =
if pos < 2*n
then open <|> close
else stop where
-- Add an open bracket if we can
open =
if depth < 2*n - pos
then ('(' :) <$> 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)
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("}"));
}
}