0921. Minimum Add to Make Parentheses Valid

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:

  1. S.length <= 1000
  2. 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)
}


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