856. Score of Parentheses
https://leetcode.com/problems/score-of-parentheses/
solution
# 左括号更新深度,右括号更新结果
class Solution:
def scoreOfParentheses(self, s: str) -> int:
score = 0
depth = 0
for i, char in enumerate(s):
if char == '(':
depth += 1
else:
depth -= 1
if s[i-1] == '(':
score += 2 ** depth
return score
时间复杂度:O() 空间复杂度:O()
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