打印n对括号的所有有效组合

Question

左：只要我们没有用完所有左括号，就可以随时插入左括号。 正确：只要不引起语法错误，我们就可以插入正确的括号。 什么时候会出现语法错误

public class parentheses {

    public static void printPar(int l, int r, char[] str, int count){  //Use recursion method to 
                                                                       // print the parentheses
        if(l == 0 && r == 0){     //if there are no parentheses available, print them out  
            System.out.println(str); //Print out the parentheses
        }

        else{
            if(l > 0){    // try a left paren, if there are some available
                str[count] = '(';
                printPar(l - 1, r, str, count + 1); //Recursion
            }
            if(r > 0){   // try a right paren, if there are some available
                str[count] = ')';
                printPar(l, r - 1, str, count + 1);  //Recursion
            }
        }
    }

    public static void printPar(int count){
        char[] str = new char[count*2];   // Create a char array to store the parentheses
        printPar(count,count,str,0);      //call the printPar method, the parameters are the left,
                                            //the right parentheses, the array to store the 
                                            //parenthese, and the counter
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        printPar(2);  //

    }

}

结果应为：

(())
()()

但是我得到的是：

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

Answer 1

import java.util.ArrayList;
import java.util.List;


public class parentheses {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println(prent(2));
    }

    public static List<String> prent(int n) {
        List<String> l = new ArrayList<String>();
        if(n==1) {
            l.add("()");
            return l;
        }
        for(String st : prent(n-1)) {
            l.add(st+"()");
            l.add("("+st+")");
            if(!(st+"()").equals("()"+st)) {
                l.add("()"+st);
            }
        }
        return l;
    }

}

Answer 2

试试这个代码。 经过测试，工作正常。

 public class ParanthesisCombination {
     public static void main(String[] args) {
          printParenthesis(3);
     }
     static void printParenthesis(int n){
        printParenthesis("",n,n);       
     }

     static void printParenthesis(String s,int open,int close){
         if(open>close)
            return;
         if(open == 0 && close == 0){
             System.out.println(s);
             return;
         }
         if(open < 0 || close<0)
             return;

         printParenthesis(s + '{',open-1,close);
         printParenthesis(s + '}',open,close-1);
     }
}

Answer 3

    private static void printA(int open, int close, int max, String out) {
        if(open==close && close==max){
            st.add(out);
            System.out.println(out);
        } else {
            if(open+1<=max){
                printA(open+1, close, max, out+"(");
            }
            if(open>close && close+1<=max){
                printA(open, close+1, max, out+")");
            }
        }
    }
    public static  ArrayList<String>st = new ArrayList<String>();
    public static void main(String[] args) {
        //2 is maximum open/close parenthese
        //i save the output in st(arraylist), 
        printA(0,0,2,"");
    }

Answer 4

我认为问题在于代码中没有任何部分执行“ no sintax error ”部分。

您应该将if(r > 0){...} if(r > l){...}更改为if(r > 0){...} if(r > l){...}以执行该规则； 如果至少有1个左括号仍未“打开”，则不应打印右括号。

Answer 5

让我们假设给出的总对数为3-要保持有效，您只需要在开始时打开大括号，在结束时一个关闭大括号，这意味着（2对组合）。

因此2个对中的总有效+无效组合= 3 * 2 =6。这些没有重复。.认为'（'为0，'）'为1。

0000 1000 0100 1100 0010 1010 0110 1110 0001 1001 0101 1101 0011 1011 0111 1111

只能有6个可能的组合具有2 0和2 1。.因此，对于2对总组合= 6因此对于3对总有效组合= 6

Answer 6

public class parentheses {
    public static void printPar(int l, int r, char[] str, int count){  //Use recursion method to 
                                                               // print the parentheses

    if(l == 0 && r == 0){     //if there are no parentheses available, print them out  
        System.out.println(str); //Print out the parentheses
    }

    else{
        if(l > 0){    // try a left paren, if there are some available
            str[count] = '(';
            printPar(l - 1, r, str, count + 1); //Recursion
        }
        // Add constraint that a parenthesis is open before inserting a closing paranthesis
        // hence, add l < r - meaning one 'l' is printed before going for an 'r'
        if(r > 0 && l < r){   // try a right paren, if there are some available
            str[count] = ')';
            printPar(l, r - 1, str, count + 1);  //Recursion
        }
    }
}

    public static void printPar(int count){
        char[] str = new char[count*2];   // Create a char array to store the parentheses
        printPar(count,count,str,0);      //call the printPar method, the parameters are the left,
                                    //the right parentheses, the array to store the 
                                    //parenthese, and the counter
    }
    public static void main(String[] args) {
        printPar(2);  //
    }

}

Answer 7

public class parentheses {
    public static void printPar(int l, int r, char[] str, int count){
          if(i<0 || r<l) return;
          ...
    }
}

您只失去了结束条件。 希望这会有所帮助。

Answer 8

public class ValidParenthesisCombi {
    public static void main(String[] args) {
        final int noOfBraces = 3;
        print(noOfBraces);
    }

    public static void print(int n){
        print("",n,n);
    }
    public static void print( String s, int open, int close){           
        if(open > close){
            return;
        }
        if((open==0)&&(close==0)){
            System.out.println(s);
        }else{
            if(open > 0){
                print(s+"{",open-1, close);
            }if(close > 0){
                print(s+"}",open, close-1);
            }
        }
    }
}