索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github:
https://github.com/illuz/leetcode
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：
- 用 DFS 可以很快做出来，能加’(‘就加’(‘，能加’)’就加’)’。（下面的 C++ 实现）
- 还有很机灵方法写出很短的 DFS 。 (Java 实现)
- 对 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)