22. Generate Parentheses
22. Generate Parentheses - Medium
Description
22. Generate Parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Example 1:
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
这道题就是一道很经典的回溯类型的题，应用回溯解法模版即可。
Solution
因为是要生成有效的括号，所以(必须在第一个，而且)必须是跟在(的后面，这样我们加两个简单的判断就可以了。
Scala 需要引入 scala.collection.mutable.ListBuffer
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33 | class Solution {
char[] chArr = new char[] {'(', ')'};
int num;
List<String> ans = new ArrayList<>();
public List<String> generateParenthesis(int n) {
if (n == 1) return List.of("()");
num = n;
backtrack(1, 0, new StringBuilder("("));
return ans;
}
private void backtrack(int left, int right, StringBuilder sb) {
if (sb.length() == 2 * num) {
ans.add(sb.toString());
return;
}
for (char c : chArr) {
if (c == '(') {
if (left < num) {
sb.append(c);
backtrack(left + 1, right, sb);
sb.deleteCharAt(sb.length() - 1);
}
} else {
if (right < left) {
sb.append(c);
backtrack(left, right + 1, sb);
sb.deleteCharAt(sb.length() - 1);
}
}
}
}
}
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29 | import scala.collection.mutable.ListBuffer
object Solution {
def generateParenthesis(n: Int): List[String] = {
val ans = new ListBuffer[String]
def backtrack(left: Int, right: Int, str: StringBuilder): Unit = {
if (str.length == 2 * n) {
ans.addOne(str.toString())
return
}
for (c <- Array('(', ')')) {
if (c == '(') {
if (left < n) {
str.append(c)
backtrack(left + 1, right, str)
str.deleteCharAt(str.length() - 1)
}
} else {
if (right < left) {
str.append(c)
backtrack(left, right + 1, str)
str.deleteCharAt(str.length() - 1)
}
}
}
}
backtrack(1, 0, new StringBuilder().append("("))
ans.toList
}
}
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