921. Minimum Add to Make Parentheses Valid #
题目 #
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 with B), where A and B are valid strings, or
- It can be written as (A), where A 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.
题目大意 #
给一个括号的字符串,如果能在这个括号字符串中的任意位置添加括号,问能使得这串字符串都能完美匹配的最少添加数是多少。
解题思路 #
这题也是栈的题目,利用栈进行括号匹配。最后栈里剩下几个括号,就是最少需要添加的数目。
代码 #
package leetcode
func minAddToMakeValid(S string) int {
if len(S) == 0 {
return 0
}
stack := make([]rune, 0)
for _, v := range S {
if v == '(' {
stack = append(stack, v)
} else if (v == ')') && len(stack) > 0 && stack[len(stack)-1] == '(' {
stack = stack[:len(stack)-1]
} else {
stack = append(stack, v)
}
}
return len(stack)
}