[LeetCode] 022. Generate Parentheses (Medium) (C++/Java/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

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022.Generate_Parentheses (Medium)

链接：

题目：https://oj.leetcode.com/problems/generate-parentheses/
代码(github)：https://github.com/illuz/leetcode

题意：

产生有 n 对括号的所有有效字符串。

分析：

1. 用 DFS 可以很快做出来，能加’(‘就加’(‘，能加’)’就加’)’。（下面的 C++ 实现）
2. 还有很机智方法写出很短的 DFS 。 (Java 实现)
3. 对于 DFS 都可以进行记忆化，用空间换时间。 (Python 实现)

代码：

C++:

class Solution {private:string tmp;void dfs(vector<string> &v, int pos, int n, int used) {if (pos == n * 2) {cout << tmp << endl;v.push_back(tmp);return;}if (used < n) {tmp.push_back('(');dfs(v, pos + 1, n, used + 1);tmp.pop_back();}if (used * 2 > pos) {tmp.push_back(')');dfs(v, pos + 1, n, used);tmp.pop_back();}}public:    vector<string> generateParenthesis(int n) {vector<string> res;if (n == 0)return res;tmp = "";dfs(res, 0, n, 0);return res;    }};

Java:

public class Solution {    public List<String> generateParenthesis(int n) {        List<String> ret = new ArrayList<String>(), inner, outter;        if (n == 0) {            ret.add("");            return ret;        }        if (n == 1) {            ret.add("()");            return ret;        }        for (int i = 0; i < n; ++i) {            inner = generateParenthesis(i);            outter = generateParenthesis(n - i - 1);            for (int j = 0; j < inner.size(); ++j) {                for (int k = 0; k < outter.size(); ++k) {                    ret.add("(" + inner.get(j) + ")" + outter.get(k));                }            }        }        return ret;    }}

Python:

class Solution:    # @param an integer    # @return a list of string    def generateParenthesis(self, n):        dp = {0: [""], 1: ["()"]}        def memorial_dfs(n):            if n not in dp:                dp[n] = []                for i in range(n):                    for inner in memorial_dfs(i):                        for outter in memorial_dfs(n - i - 1):                            dp[n].append('(' + inner + ')' + outter)            return dp[n]        return memorial_dfs(n)