[leetcode]Generate Parentheses

简单题。DFS。还是用Java写函数麻烦了。一开始把left<n的条件忘了。

import java.util.ArrayList;

public class Solution {
    public ArrayList<String> generateParenthesis(int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<String> arr = new ArrayList<String>();
    	ArrayList<Character> chars = new ArrayList<Character>();
    	dfs(arr, 0, 0, chars, n);
    	return arr;
    }
    
    private void dfs(ArrayList<String> arr, int left, int right, ArrayList<Character> chars, int n)
    {
    	if (left + right + 1 == n * 2)
    	{
    		if (left < right) {
    			// chars.add('(')  - impossible
    			// chars.remove(chars.size() - 1);
    		}
    		else
    		{
    			chars.add(')');
    			StringBuilder sb = new StringBuilder();
    			for (char c : chars) {
    				sb.append(c);
    			}
    			arr.add(sb.toString());
    			chars.remove(chars.size() - 1);
    		}
    	}
    	else
    	{
    		if (left > right) {
    			chars.add(')');
    			dfs(arr, left, right + 1, chars, n);
    			chars.remove(chars.size() - 1);
    		}
    		if (left < n)
    		{
	    		chars.add('(');
	    		dfs(arr, left + 1, right, chars, n);
	    		chars.remove(chars.size() - 1);
    		}
    	}
    }
}

Python3

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        output = []
        def backtrack(s: str, leftCnt: int, rightCnt: int, n: int):
            if n == leftCnt and n == rightCnt:
                output.append(s)
            else:
                if leftCnt < n:
                    backtrack(s + '(', leftCnt + 1, rightCnt, n)
                if rightCnt < leftCnt:
                    backtrack(s + ')', leftCnt, rightCnt + 1, n)
        if n > 0:
            backtrack('', 0, 0, n)
        return output