Minimum Add to Make Parentheses Valid
阿新 • • 發佈:2018-12-25
Given a string S
of '('
and ')'
parentheses, we add the minimum number of parentheses ( '('
or ')'
, and in any positions ) so that the resulting parentheses string is valid.
Formally, a parentheses string is valid if and only if:
- It is the empty string, 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.
Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.
Example 1:
Input: "())" Output: 1
Example 2:
Input: "(((" Output: 3
Example 3:
Input: "()" Output: 0
Example 4:
Input: "()))((" Output: 4
Note:
S.length <= 1000
S
only consists of'('
and')'
characters.
題目理解:
給定一個只含有左括號和右括號的陣列,判斷還需要新增多少個括號才能時得括號合法
解題思路:
分別找出無法匹配的左括號和無法匹配的右括號就可以了,注意")("這種情況是不能匹配的
程式碼如下:
class Solution {
public int minAddToMakeValid(String S) {
int a = 0, b = 0;
for(char ch : S.toCharArray()){
if(ch == '(')
a++;
else{
if(a > 0)
a--;
else
b++;
}
}
return a + b;
}
}