LeetCode 22 — Generate Parentheses（C++ Java Python）

题目： http://oj.leetcode.com/problems/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:

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

题目翻译：

给定n对括号，写一个函数来生成成对的括号的所有组合。
例如，给定n = 3，一组方案集是：
"((()))", "(()())", "(())()", "()(())", "()()()"

分析：
        递归实现，如果左括号还有剩余，则可以放置左括号，如果右括号的剩余数大于左括号，则可以放置右括号。
C++实现：

class Solution {
public:
    vector generateParenthesis(int n) {
        vector res;
    	generate(n, n, "", res);
    	return res;
    }

    void generate(int left, int right, string str, vector& res)
    {
    	if(left == 0 && right == 0)
    	{
    		res.push_back(str);
    		return;
    	}
    	if(left > 0)
    	{
    		generate(left - 1, right, str + '(', res);
    	}
    	if(right > left)
    	{
    		generate(left, right - 1, str + ')', res);
    	}
    }
};

Java实现：

public class Solution {
    public ArrayList generateParenthesis(int n) {
        ArrayList res = new ArrayList();
		generate(n, n, "", res);
		return res;
	}

	public void generate(int left, int right, String str, ArrayList res) {
		if (left == 0 && right == 0) {
			res.add(str);
			return;
		}
		if (left > 0) {
			generate(left - 1, right, str + '(', res);
		}
		if (right > left) {
			generate(left, right - 1, str + ')', res);
		}
    }
}

Python实现：

class Solution:
    # @param an integer
    # @return a list of string
    def generateParenthesis(self, n):
        res = []
        self.generate(n, n, "", res)
        return res
    
    def generate(self, left, right, str, res):
        if left == 0 and right == 0:
            res.append(str)
            return
        if left > 0:
            self.generate(left - 1, right, str + '(', res)
        if right > left:
            self.generate(left, right - 1, str + ')', res)

        感谢阅读，欢迎评论！