1249. Minimum Remove to Make Valid Parentheses (M)
Minimum Remove to Make Valid Parentheses (M)
题目
Given a string s of '('
, ')'
and lowercase English characters.
Your task is to remove the minimum number of parentheses ( '('
or ')'
, in any positions ) so that the resulting parentheses string is valid and return any valid string.
Formally, a parentheses string is valid if and only if:
- It is the empty string, contains only lowercase characters, or
- It can be written as
AB
(A
concatenated withB
), whereA
andB
are valid strings, or - It can be written as
(A)
, whereA
is a valid string.
Example 1:
Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
Example 2:
Input: s = "a)b(c)d"
Output: "ab(c)d"
Example 3:
Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.
Example 4:
Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"
Constraints:
1 <= s.length <= 10^5
s[i]
is one of'('
,')'
and lowercase English letters.
题意
将给定字符串中非法的括号删去。
思路
只要保证左右括号数量相等即可。先从左到右遍历,将所有无法匹配的')'删去,这样得到的字符串保证了'('的数量一定大于等于')'的数量;再从右到左遍历新字符串,将所有无法匹配的'('删去,最终得到的字符串一定满足'('和')'数量相等。
代码实现
Java
class Solution {
public String minRemoveToMakeValid(String s) {
StringBuilder sb = new StringBuilder();
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c != ')') {
sb.append(c);
if (c == '(') cnt++;
} else if (cnt > 0) {
cnt--;
sb.append(c);
}
}
if (cnt == 0) return sb.toString();
s = sb.toString();
sb = new StringBuilder();
cnt = 0;
for (int i = s.length() - 1; i >= 0; i--) {
char c = s.charAt(i);
if (c != '(') {
sb.insert(0, c);
if (c == ')') cnt++;
} else if (cnt > 0) {
cnt--;
sb.insert(0, c);
}
}
return sb.toString();
}
}