https://leetcode.com/problems/valid-parentheses/
Algorithm
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> FUNCTION main(): O(n), n = length of input string
- DO Read user input and return problem output
- INPUT s (input string): string
- OUTPUT validness of INPUT: bool
GLOBAL st ← empty stack
FOR each character c, index i in s
CALL validness ← check(c, i)
IF validness is False THEN
return False
IF st is not empty THEN
return False
return True
---
> FUNCTION check(): O(1)
- DO Read a character and return if it's valid under some conditions
- INPUT c (checked character): character, i (index of character): integer
- OUTPUT validness of INPUT: bool
CALL c ← convert(c)
IF st is empty THEN
IF c % 2 = 0 THEN
return False
ELSE
push c to st
return True
top ← top element of st
IF c - top != 1 THEN
IF c % 2 = 0 THEN
return False
ELSE
push c to st
return True
ELSE
pop st
return True
---
> FUNCTION convert(): O(1)
- DO Read a character and return a converted integer
- INPUT c: character
- OUTPUT converted integer: integer
CASE c OF
( : return 1
) : return 2
{ : return 3
} : return 4
[ : return 5
] : return 6
exit with error code -1 (Input string includes unexpected character)
Code
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class Solution:
def __init__(self):
self.st = []
def isValid(self, s: str) -> bool:
for i, c in enumerate(s):
if not self.check(c, i):
return False
if len(self.st) != 0:
return False
return True
def check(self, c, i):
c = self.convert(c)
if len(self.st) == 0:
if c % 2 == 0: # )
return False
else: # (
self.st.append(c)
return True
top = self.st[-1] # 1
if c - top != 1:
if c % 2 == 0: # ()
return False
else: # ()
self.st.append(c) # 1
return True
else:
self.st.pop()
return True
def convert(self, c):
if c == '(':
return 1
elif c == ')':
return 2
elif c == '{':
return 3
elif c == '}':
return 4
elif c == '[':
return 5
elif c == ']':
return 6
exit(-1)
Runtime: 24 ms, faster than 89.69% of Python3 online submissions for Valid Parentheses.
Memory Usage: 12.9 MB, less than 100.00% of Python3 online submissions for Valid Parentheses.
55분 소요
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