[42] (Backtracking method) Bracket generation (LC 22)

Bracket generation

Problem Description

The number n represents the logarithm of generating brackets. Please design a function to generate all possible and effective bracket combinations.

example:

Problem-solving ideas

Use the backtracking method to exhaustively, use the One_of_ans[2*n] array to store the result, if One_of_ans[i] is 0, it is the right parenthesis, if One_of_ans[i] is 1, it is the left parenthesis, traverse the subset tree to get a feasible solution, Then convert the One_of_ans[] into a string to get the answer to this question.

Code

class Solution {
    
    
    public List<String> generateParenthesis(int n) {
    
    
        List<String> res = new ArrayList<String>();
        int One_of_ans[] = new int[2*n];
        backtrack(res,One_of_ans,2*n,0);
        return res;
    }

    public void backtrack(List<String> res,int One_of_ans[],int n,int t){
    
    //回溯
     //   System.out.print("现在在第"+t+"层：");
        if(t>=n)
            output(res,One_of_ans,n);
        else{
    
    
            for(int i=0;i<=1;i++){
    
    
                One_of_ans[t] = i;
                backtrack(res,One_of_ans,n,t+1);
            }
        }
    }

    public void output(List<String> res,int One_of_ans[],int n){
    
     //根据整型数组得到结果
        if(legal(One_of_ans,n)){
    
    
            String OneOfAns = "";
            for(int i : One_of_ans){
    
    
                if(i == 0)
                    OneOfAns = OneOfAns +"(";
                else
                    OneOfAns = OneOfAns+ ")";
            }
            res.add(OneOfAns);
        }
    }

    public boolean legal(int One_of_ans[],int n){
    
    //判断是否为合法括号
        for(int i : One_of_ans)
            System.out.print(i);
        System.out.println();
        Stack<Integer> s = new Stack<>();
        for(int i : One_of_ans){
    
    
            if(i == 0)
                s.push(new Integer(1));
            else if(s.isEmpty() || !s.pop().equals((Integer)i))
                return false;
        }
        if(!s.isEmpty())
            return false;
        return true;
    }
}

Small summary

Note on the Java grammar obtained from this question:

1. The parameter of Java's Stack class is the object type;
2. The comparison of object types uses equals() method instead of ==.