0856. Score of Parentheses

856. Score of Parentheses #

题目 #

Given a balanced parentheses string S, compute the score of the string based on the following rule:

() has score 1 AB has score A + B, where A and B are balanced parentheses strings. (A) has score 2 * A, where A is a balanced parentheses string.

Example 1:


Input: "()"
Output: 1

Example 2:


Input: "(())"
Output: 2

Example 3:


Input: "()()"
Output: 2

Example 4:


Input: "(()(()))"
Output: 6

Note:

  1. S is a balanced parentheses string, containing only ( and ).
  2. 2 <= S.length <= 50

题目大意 #

按照以下规则计算括号的分数:() 代表 1 分。AB 代表 A + B,A 和 B 分别是已经满足匹配规则的括号组。(A) 代表 2 * A,其中 A 也是已经满足匹配规则的括号组。给出一个括号字符串,要求按照这些规则计算出括号的分数值。

解题思路 #

按照括号匹配的原则,一步步的计算每个组合的分数入栈。遇到题目中的 3 种情况,取出栈顶元素算分数。

代码 #


package leetcode

func scoreOfParentheses(S string) int {
	res, stack, top, temp := 0, []int{}, -1, 0
	for _, s := range S {
		if s == '(' {
			stack = append(stack, -1)
			top++
		} else {
			temp = 0
			for stack[top] != -1 {
				temp += stack[top]
				stack = stack[:len(stack)-1]
				top--
			}
			stack = stack[:len(stack)-1]
			top--
			if temp == 0 {
				stack = append(stack, 1)
				top++
			} else {
				stack = append(stack, temp*2)
				top++
			}
		}
	}
	for len(stack) != 0 {
		res += stack[top]
		stack = stack[:len(stack)-1]
		top--
	}
	return res
}


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