• [leetcode] 22. Generate Parentheses


    题目大意

    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:

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

    解题思路

    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.

    伪代码(递归法)

    if len(s) == 2*n:
        将s加入到输出列表
    if 左括号数目 < n:
        加入左括号
    if 右括号数目 < 左括号数目:
        加入右括号

    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

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

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  • 原文地址:https://www.cnblogs.com/bymo/p/9644896.html
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