[LeetCode] 1021. Remove Outermost Parentheses
grandyang opened this issue · comments
A valid parentheses string is either empty ("")
, "(" + A + ")"
, or A + B
, where A
and B
are valid parentheses strings, and +
represents string concatenation. For example, ""
, "()"
, "(())()"
, and "(()(()))"
are all valid parentheses strings.
A valid parentheses string S
is primitive if it is nonempty, and there does not exist a way to split it into S = A+B
, with A
and B
nonempty valid parentheses strings.
Given a valid parentheses string S
, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k
, where P_i
are primitive valid parentheses strings.
Return S
after removing the outermost parentheses of every primitive string in the primitive decomposition of S
.
Example 1:
Input: "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
Note:
S.length <= 10000
S[i]
is"("
or")"
S
is a valid parentheses string
这道题给了一个合法的括号字符串,其可能由多个合法的括号字符子串组成,现在让把所有合法的子串的最外层的括号去掉,将剩下的拼接起来并返回,根据题目给的例子,不难理解题意。LeetCode 中关于括号的题目还是比较多的,比如 Valid Parentheses,Valid Parenthesis String,Remove Invalid Parentheses,和 Longest Valid Parentheses 等。大多都是考察如何判断一个括号字符串是否合法,所谓的合法,大致就是左右括号个数要相同,每个右括号前面必须要有对应的左括号,一个比较简单的判断方法就是用一个变量 cnt,遇到左括号则自增1,遇到右括号则自减1,在这过程中 cnt 不能为负,且最后 cnt 必须为0。这道题限定了括号字符串一定是合法的,但也可以用这个方法来找出每个合法的子串部分,遍历字符串S,若当前字符为左括号,则 cnt 自增1,否则自减1。若 cnt 不为0,说明还不是一个合法的括号子串,跳过。否则我们就知道了一个合法括号子串的结束位置,用一个变量 start 记录合法括号子串的起始位置,初始化为0,这样就可以将去除最外层括号后的中间部分直接取出来加入结果 res 中,然后此时更新 start 为下一个合法子串的起始位置继续遍历即可,参见代码如下:
解法一:
class Solution {
public:
string removeOuterParentheses(string S) {
string res = "";
int cnt = 0, start = 0, n = S.size();
for (int i = 0; i < n; ++i) {
(S[i] == '(') ? ++cnt : --cnt;
if (cnt != 0) continue;
res += S.substr(start + 1, i - start - 1);
start = i + 1;
}
return res;
}
};
我们也可以写的更简洁一些,并不需要等到找到整个合法括号子串后再加入结果 res,而是在遍历的过程中就加入。因为这里的括号分为两种,一种是合法子串的最外层括号,这种不能加到结果 res,另一种是其他位置上的括号,这种要加到 res。所以只要区分出这两种情况,就知道当前括号要不要加,区别的方法还是根据 cnt,当遇到左括号时,若此时 cnt 大于0,则一定不是合法子串的起始位置,可以加入 res,之后 cnt 自增1;同理,若遇到右括号,若此时 cnt 大于1,则一定不是合法子串的结束位置,可以加入 res,之后 cnt 自减1,参见代码如下:
解法二:
class Solution {
public:
string removeOuterParentheses(string S) {
string res;
int cnt = 0;
for (char c : S) {
if (c == '(' && cnt++ > 0) res.push_back(c);
if (c == ')' && cnt-- > 1) res.push_back(c);
}
return res;
}
};
Github 同步地址:
类似题目:
参考资料:
public String removeOuterParentheses(String S) {
char[] chars = S.toCharArray();
int level = 0;
int length = S.length();
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < length; i++) {
if (chars[i] == '(') {
if (level > 0) {
stringBuilder.append(chars[i]);
}
level++;
} else {
level--;
if (level > 0) {
stringBuilder.append(chars[i]);
}
}
}
return stringBuilder.toString();
}