[LeetCode] 22. Generate Parentheses 生成括号

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:

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

给一个数字n，生成所有长度为2n配对正确的括号组合。

使用递归Recursion一步步构造字符串，当左括号出现次数<n时，就可以放置新的左括号。当右括号出现次数小于左括号出现次数时，就可以放置新的右括号。

Java:

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<>();
        StringBuilder path = new StringBuilder();
        if (n > 0) generate(n, path, result, 0, 0);
        return result;
    }
    // l 表示 ‘( ‘出现的次数, r 表示 ‘)‘ 出现的次数
    private static void generate(int n, StringBuilder path,
                                 List<String> result, int l, int r) {
        if (l == n) {
            StringBuilder sb = new StringBuilder(path);
            for (int i = 0; i < n - r; ++i) sb.append(‘)‘);
            result.add(sb.toString());
            return;
        }

        path.append(‘(‘);
        generate(n, path, result, l + 1, r);
        path.deleteCharAt(path.length() - 1);

        if (l > r) {
            path.append(‘)‘);
            generate(n, path, result, l, r + 1);
            path.deleteCharAt(path.length() - 1);
        }
    }
}

Python:

class Solution:
    # @param an integer
    # @return a list of string
    def generateParenthesis(self, n):
        result = []
        self.generateParenthesisRecu(result, "", n, n)
        return result

    def generateParenthesisRecu(self, result, current, left, right):
        if left == 0 and right == 0:
            result.append(current)
        if left > 0:
            self.generateParenthesisRecu(result, current + "(", left - 1, right)
        if left < right:
            self.generateParenthesisRecu(result, current + ")", left, right - 1)

C++:

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        generateParenthesisDFS(n, n, "", res);
        return res;
    }
    void generateParenthesisDFS(int left, int right, string out, vector<string> &res) {
        if (left > right) return;
        if (left == 0 && right == 0) res.push_back(out);
        else {
            if (left > 0) generateParenthesisDFS(left - 1, right, out + ‘(‘, res);
            if (right > 0) generateParenthesisDFS(left, right - 1, out + ‘)‘, res);
        }
    }
};

类似题目：

[LeetCode] 20. Valid Parentheses 合法括号

[LeetCode] 32. Longest Valid Parentheses 最长有效括号

原文地址：https://www.cnblogs.com/lightwindy/p/8497012.html