【LeetCode】Generate Parentheses (2 solutions)

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:

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

解法一：递归

借助栈，‘(‘、‘)‘构成一对分别进出栈。最后栈为空，则输入括号构成的字符串是合法的。

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        if(n == 0)
            return result;
        //first must be ‘(‘
        string cur = "(";
        stack<char> s;
        s.push(‘(‘);

        Helper(result, cur, s, 2*n-1);
        return result;
    }
    void Helper(vector<string>& result, string cur, stack<char> s, int num)
    {
        if(num == 1)
        {//must be ‘)‘
            if(s.top() == ‘(‘ && s.size() == 1)
            {//all matched
                cur += ‘)‘;
                result.push_back(cur);
            }
        }
        else
        {
            //‘(‘ always push
            string str1 = cur;
            str1 += ‘(‘;
            s.push(‘(‘);
            Helper(result, str1, s, num-1);
            s.pop();

            //‘)‘
            if(!s.empty())
            {//prune. never begin with ‘)‘
                string str2 = cur;
                str2 += ‘)‘;
                if(s.top() == ‘(‘)
                    s.pop();    //check empty() before access top()
                else
                    s.push(‘)‘);
                Helper(result, str2, s, num-1);
            }
        }
    }
};

解法二：递归

稍作分析可知，栈是不必要的，只要记录字符串中有几个‘(‘，记为count。

每进入一个‘(‘, count ++. 每匹配一对括号, count--。

最终全部匹配，需要count==0

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        if(n == 0)
            return result;
        //first must be ‘(‘
        string cur = "(";
        int count = 1;  //number of (‘s in cur

        Helper(result, cur, count, 2*n-1);
        return result;
    }
    void Helper(vector<string>& result, string cur, int count, int num)
    {
        if(num == 1)
        {//must be ‘)‘
            if(count == 1)
            {//all matched
                cur += ‘)‘;
                result.push_back(cur);
            }
        }
        else
        {
            //‘(‘ always push
            string str1 = cur;
            str1 += ‘(‘;
            count ++;
            Helper(result, str1, count, num-1);
            count --;

            //‘)‘
            if(count != 0)
            {//prune. never begin with ‘)‘
                string str2 = cur;
                str2 += ‘)‘;
                count --;
                Helper(result, str2, count, num-1);
            }
        }
    }
};