Generate Parentheses 解答

Question

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()

Solution 1

Two conditions to check:

if remained left '(' nums < remained right ')', for next char, we can put '(' or ')'.

if remained left '(' nums = remained right ')', for next char, we can only put '('.

Draw out the solution tree, and do DFS.

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<String>();
        List<Character> list = new ArrayList<Character>();
        dfs(n, n, list, result);
        return result;
    }
    
    private void dfs(int leftRemain, int rightRemain, List<Character> list, List<String> result) {
        if (leftRemain < 0 || rightRemain < 0 || leftRemain > rightRemain)
            return;
        
        if (leftRemain == 0 && rightRemain == 0) {
            int size = list.size();
            StringBuilder sb = new StringBuilder(size);
            for (char tmpChar : list)
                sb.append(tmpChar);
            result.add(sb.toString());
            return;
        }
        
        if (leftRemain == rightRemain) {
            list.add('(');
            dfs(leftRemain - 1, rightRemain, list, result);
            list.remove(list.size() - 1);
        } else {
            list.add(')');
            dfs(leftRemain, rightRemain - 1, list, result);
            list.remove(list.size() - 1);
            list.add('(');
            dfs(leftRemain - 1, rightRemain, list, result);
            list.remove(list.size() - 1);
        }
    }
}

Solution 2

A much simpler solution.

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<String>();
        List<Character> list = new ArrayList<Character>();
        dfs(n, n, "", result);
        return result;
    }
    
    private void dfs(int leftRemain, int rightRemain, String prefix, List<String> result) {
        if (leftRemain < 0 || rightRemain < 0 || leftRemain > rightRemain)
            return;
        if (leftRemain == 0 && rightRemain == 0) {
            result.add(new String(prefix));
            return;
        }
        if (leftRemain > 0)
            dfs(leftRemain - 1, rightRemain, prefix + "(", result);
        if (rightRemain > 0)
            dfs(leftRemain, rightRemain - 1, prefix + ")", result);
    }
}