22. Generate Parentheses

本文介绍了一个使用递归方法生成所有合法括号组合的算法。针对输入的整数n,该算法能够生成所有可能的由n对括号组成的合法字符串组合。文章通过具体的代码示例详细解释了递归过程及实现细节。

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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:

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


public class Solution{
	public void solution(List<String> res, String s, int left, int right) {
        if(left!=0){
            solution(res, s + "(", left-1, right);
            if(left < right)
                solution(res, s + ")", left, right-1);
        //left!=0,无脑递归
        }else{
            while(right!=0){
                s = s + ")";
                right--;
            }
            res.add(s);
        //这种情况下就是把剩下的")"补上,因为题主不会直接添加指定长度重复的字符串,
        //所以只能用笨方法。这里优化的话,能打败跟自己同运行时间的,大概60%的人
        }
        return;
    }
    
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        solution(res, "", n, n);
        return res;
    }
}

#include <cassert> /// for assert #include <iostream> /// for I/O operation #include <vector> /// for vector container /** @brief Backtracking algorithms @namespace backtracking / namespace backtracking { /* @brief generate_parentheses class */ class generate_parentheses { private: std::vectorstd::string res; ///< Contains all possible valid patterns void makeStrings(std::string str, int n, int closed, int open); public: std::vectorstd::string generate(int n); }; /** @brief function that adds parenthesis to the string. @param str string build during backtracking @param n number of pairs of parentheses @param closed number of closed parentheses @param open number of open parentheses */ void generate_parentheses::makeStrings(std::string str, int n, int closed, int open) { if (closed > open) // We can never have more closed than open return; if ((str.length() == 2 * n) && (closed != open)) { // closed and open must be the same return; } if (str.length() == 2 * n) { res.push_back(str); return; } makeStrings(str + ')', n, closed + 1, open); makeStrings(str + '(', n, closed, open + 1); } /** @brief wrapper interface @param n number of pairs of parentheses @return all well-formed pattern of parentheses */ std::vectorstd::string generate_parentheses::generate(int n) { backtracking::generate_parentheses::res.clear(); std::string str = “(”; generate_parentheses::makeStrings(str, n, 0, 1); return res; } } // namespace backtracking /** @brief Self-test implementations @returns void */ static void test() { int n = 0; std::vectorstd::string patterns; backtracking::generate_parentheses p; n = 1; patterns = {{“()”}}; assert(p.generate(n) == patterns); n = 3; patterns = {{“()()()”}, {“()(())”}, {“(())()”}, {“(()())”}, {“((()))”}}; assert(p.generate(n) == patterns); n = 4; patterns = {{“()()()()”}, {“()()(())”}, {“()(())()”}, {“()(()())”}, {“()((()))”}, {“(())()()”}, {“(())(())”}, {“(()())()”}, {“(()()())”}, {“(()(()))”}, {“((()))()”}, {“((())())”}, {“((()()))”}, {“(((())))”}}; assert(p.generate(n) == patterns); std::cout << “All tests passed\n”; } /** @brief Main function @returns 0 on exit */ int main() { test(); // run self-test implementations return 0; } 在这段代码的基础上为C++初学者出几个练习题?
最新发布
03-08
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