LeetCode之Minimum Add to Make Parentheses Valid（Kotlin）

問題： 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.
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方法： 像這種左右對稱的算法題主要經過棧去處理，遇到成對的"()"就從棧中pop，最後剩下的就是沒有成對的，也就是題目要計算的數量。

具體實現：

import java.util.*

class MinimumAddToMakeParenthesesValid {
    fun minAddToMakeValid(S: String): Int {
        val stack = ArrayDeque<Char>()
        for (ch in S) {
            if (stack.isEmpty()) {
                stack.push(ch)
                continue
            }
            if (ch == ')' && stack.first == '(') {
                stack.pop()
            } else {
                stack.push(ch)
            }
        }
        return stack.size
    }
}

fun main(args: Array<String>) {
    val input = "())"
    val minimumAddToMakeParenthesesValid = MinimumAddToMakeParenthesesValid()
    println(minimumAddToMakeParenthesesValid.minAddToMakeValid(input))
}
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有問題隨時溝通

具體代碼實現能夠參考Github