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:
"((()))", "(()())", "(())()", "()(())", "()()()"
给一个数字n，生成所有长度为2n配对正确的括号组合。
使用递归Recursion一步步构造字符串，当左括号出现次数<n时，就可以放置新的左括号。当右括号出现次数小于左括号出现次数时，就可以放置新的右括号。
Java:
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<>();
StringBuilder path = new StringBuilder();
if (n > 0) generate(n, path, result, 0, 0);
return result;
}
// l 表示 ‘( ‘出现的次数, r 表示 ‘)‘ 出现的次数
private static void generate(int n, StringBuilder path,
List<String> result, int l, int r) {
if (l == n) {
StringBuilder sb = new StringBuilder(path);
for (int i = 0; i < n - r; ++i) sb.append(‘)‘);
result.add(sb.toString());
return;
}
path.append(‘(‘);
generate(n, path, result, l + 1, r);
path.deleteCharAt(path.length() - 1);
if (l > r) {
path.append(‘)‘);
generate(n, path, result, l, r + 1);
path.deleteCharAt(path.length() - 1);
}
}
}
Python:
class Solution:
# @param an integer
# @return a list of string
def generateParenthesis(self, n):
result = []
self.generateParenthesisRecu(result, "", n, n)
return result
def generateParenthesisRecu(self, result, current, left, right):
if left == 0 and right == 0:
result.append(current)
if left > 0:
self.generateParenthesisRecu(result, current + "(", left - 1, right)
if left < right:
self.generateParenthesisRecu(result, current + ")", left, right - 1)
C++:
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);
}
}
};
类似题目：
[LeetCode] 20. Valid Parentheses 合法括号
[LeetCode] 32. Longest Valid Parentheses 最长有效括号
