[LeetCode 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:

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

Diffculty Easy

Similar Problems [LeetCode ] Letter Combinations of a Phone Number Medium [LeetCode ] Valid Parentheses Easy

Analysis

通过不断插入 "(" 和 ")" 直到两者的数量都为 n ，则一个 combination 构建完成（递归终止条件）。如何保证这个 combination 是 well-formed ？在插入过程中的任何时候：

1、只要 "(" 的数量没有超过 n，都可以插入 "("。 2、而可以插入 ")" 的前提则是当前的 "(" 数量必须要多余当前的 ")" 数量。

Solutions

1、递归

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> allComb;
        string comb;
        findParenthesis(n, 0, 0, comb, allComb);
        return allComb;
    }

    void findParenthesis(int n, int nleft, int nright, string &comb, vector<string> &allComb) {
        if(nleft == n && nright == n) { // comb.size() == 2 * n
            allComb.push_back(comb);
            return;
        }

        if(nleft < n) {
            comb.push_back('(');
            findParenthesis(n, nleft + 1, nright, comb, allComb);
            comb.pop_back();
        }

        if(nright < nleft) {
            comb.push_back(')');
            findParenthesis(n, nleft, nright + 1, comb, allComb);
            comb.pop_back();
        }
    }
};

2、递归 DFS

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);
        }
    }
};