[LeetCode] Generate Parentheses 生成括号

时间:2023-03-08 17:24:35
[LeetCode] Generate Parentheses 生成括号

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:

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

在 LeetCode 中有关括号的题共有七道,除了这一道的另外六道是 Score of ParenthesesValid Parenthesis String, Remove Invalid ParenthesesDifferent Ways to Add ParenthesesValid Parentheses 和 Longest Valid Parentheses。这道题给定一个数字n,让生成共有n个括号的所有正确的形式,对于这种列出所有结果的题首先还是考虑用递归 Recursion 来解,由于字符串只有左括号和右括号两种字符,而且最终结果必定是左括号3个,右括号3个,所以这里定义两个变量 left 和 right 分别表示剩余左右括号的个数,如果在某次递归时,左括号的个数大于右括号的个数,说明此时生成的字符串中右括号的个数大于左括号的个数,即会出现 ')(' 这样的非法串,所以这种情况直接返回,不继续处理。如果 left 和 right 都为0,则说明此时生成的字符串已有3个左括号和3个右括号,且字符串合法,则存入结果中后返回。如果以上两种情况都不满足,若此时 left 大于0,则调用递归函数,注意参数的更新,若 right 大于0,则调用递归函数,同样要更新参数,参见代码如下:

C++ 解法一:

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 == && right == ) res.push_back(out);
else {
if (left > ) generateParenthesisDFS(left - , right, out + '(', res);
if (right > ) generateParenthesisDFS(left, right - , out + ')', res);
}
}
};

Java 解法一:

public class Solution {
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<String>();
helper(n, n, "", res);
return res;
}
void helper(int left, int right, String out, List<String> res) {
if (left < 0 || right < 0 || left > right) return;
if (left == 0 && right == 0) {
res.add(out);
return;
}
helper(left - 1, right, out + "(", res);
helper(left, right - 1, out + ")", res);
}
}

再来看那一种方法,这种方法是 CareerCup 书上给的方法,感觉也是满巧妙的一种方法,这种方法的思想是找左括号,每找到一个左括号,就在其后面加一个完整的括号,最后再在开头加一个 (),就形成了所有的情况,需要注意的是,有时候会出现重复的情况,所以用set数据结构,好处是如果遇到重复项,不会加入到结果中,最后我们再把set转为vector即可,参见代码如下::

n=1:    ()

n=2:    (())    ()()

n=3:    (()())    ((()))    ()(())    (())()    ()()()

C++ 解法二:

class Solution {
public:
vector<string> generateParenthesis(int n) {
unordered_set<string> st;
if (n == ) st.insert("");
else {
vector<string> pre = generateParenthesis(n - );
for (auto a : pre) {
for (int i = ; i < a.size(); ++i) {
if (a[i] == '(') {
a.insert(a.begin() + i + , '(');
a.insert(a.begin() + i + , ')');
st.insert(a);
a.erase(a.begin() + i + , a.begin() + i + );
}
}
st.insert("()" + a);
}
}
return vector<string>(st.begin(), st.end());
}
};

Java 解法二:

public class Solution {
public List<String> generateParenthesis(int n) {
Set<String> res = new HashSet<String>();
if (n == 0) {
res.add("");
} else {
List<String> pre = generateParenthesis(n - 1);
for (String str : pre) {
for (int i = 0; i < str.length(); ++i) {
if (str.charAt(i) == '(') {
str = str.substring(0, i + 1) + "()" + str.substring(i + 1, str.length());
res.add(str);
str = str.substring(0, i + 1) + str.substring(i + 3, str.length());
}
}
res.add("()" + str);
}
}
return new ArrayList(res);
}
}

Github 同步地址:

https://github.com/grandyang/leetcode/issues/22

类似题目:

Remove Invalid Parentheses

Different Ways to Add Parentheses

Longest Valid Parentheses

Valid Parentheses

Score of Parentheses

Valid Parenthesis String

参考资料:

https://leetcode.com/problems/generate-parentheses/

https://leetcode.com/problems/generate-parentheses/discuss/10127/An-iterative-method.

https://leetcode.com/problems/generate-parentheses/discuss/10337/My-accepted-JAVA-solution

https://leetcode.com/problems/generate-parentheses/discuss/10105/Concise-recursive-C%2B%2B-solution

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