[leetcode] 22. Generate Parentheses

題目大意

https://leetcode.com/problems/generate-parentheses/description/

22. Generate Parentheses

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:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

 

解題思路

The idea here is to only add '(' and ')' that we know will guarantee us a solution (instead of adding 1 too many close). Once we add a '(' we will then discard it and try a ')' which can only close a valid '('. Each of these steps are recursively called.

僞代碼(遞歸法)

if len(s) == 2*n:
    將s加入到輸出列表
if 左括號數目 < n:
    加入左括號
if 右括號數目 < 左括號數目:
    加入右括號

 

Java解法

 public List<String> generateParenthesis(int n) {
        List<String> list = new ArrayList<String>();
        backtrack(list, "", 0, 0, n);
        return list;
    }
    
    public void backtrack(List<String> list, String str, int open, int close, int max){
        
        if(str.length() == max*2){
            list.add(str);
            return;
        }
        
        if(open < max)
            backtrack(list, str+"(", open+1, close, max);
        if(close < open)
            backtrack(list, str+")", open, close+1, max);
    }

 

Python解法

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        self.output_list = list()
        self.n = n
        
        def backtrack(s, left, right):
            if len(s) == n * 2:
                self.output_list.append(s)
                return
            if left < self.n:
                backtrack(s + "(", left + 1, right)
            if right < left:
                backtrack(s + ")", left, right + 1)
        
        backtrack("", 0, 0)
        return self.output_list

 

參考：https://leetcode.com/problems/generate-parentheses/discuss/10100/Easy-to-understand-Java-backtracking-solution