LeetCode----Generate Parentheses

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:

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

分析：

生成合法的括号串。

递归：

每次先判断当前串中的左括号数目是否大于等于右括号数目，如果成立，那么向当前子串中添加左括号或者右括号。

Python代码：

class Solution(object):    def generateParenthesis(self, n):        """        :type n: int        :rtype: List[str]        """        res = []        self.dfs('', n, res)        return res    def dfs(self, cur_s, n, res):        if len(cur_s) == 2 * n:            res.append(cur_s)            return        l_n, r_n = cur_s.count('('), cur_s.count(')')        if l_n >= r_n:            if l_n < n:                self.dfs(cur_s + '(', n, res)            if r_n < n:                self.dfs(cur_s + ')', n, res)

对应的C++代码：

class Solution {public:    vector<string> generateParenthesis(int n) {        vector<string> res;        dfs("", 0, 0, n, res);        return res;    }    void dfs(string cur_s, int l, int r, int n, vector<string> & res){        if(cur_s.length() == 2 * n){            res.push_back(cur_s);            return;        }        if(l >= r){            if(l < n){                dfs(cur_s + '(', l+1, r, n, res);            }            if(r < n){                dfs(cur_s + ')', l, r+1, n, res);            }        }    }};

别人家的代码：

Discuss中看到的动态规划：

To generate all n-pair parentheses, we can do the following:

1. Generate one pair: ()

2. Generate 0 pair inside, n - 1 afterward: () (...)...

   Generate 1 pair inside, n - 2 afterward: (()) (...)...

   ...

   Generate n - 1 pair inside, 0 afterward: ((...)) 

I bet you see the overlapping subproblems here. Here is the code:

(you could see in the code that x represents one j-pair solution and y represents one (i - j - 1) pair solution, and we are taking into account all possible of combinations of them)

class Solution(object):    def generateParenthesis(self, n):        """        :type n: int        :rtype: List[str]        """        dp = [[] for i in range(n + 1)]        dp[0].append('')        for i in range(n + 1):            for j in range(i):                dp[i] += ['(' + x + ')' + y for x in dp[j] for y in dp[i - j - 1]]        return dp[n]