Commit 38caf52

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Merge pull request cy69855522#18 from Lebhoryi/master

更新 20 valid paratheses 的暴力求解法

2 parents 0c9831d + 8cdac46 commit 38caf52Copy full SHA for 38caf52

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+10

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Lines changed: 10 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -256,6 +256,15 @@ class Solution:`
`256`	`256`	`def isValid(self, s: str) -> bool:`
`257`	`257`	`while any(('()' in s, '[]' in s, '{}' in s)): s = s.replace('()', '').replace('[]', '').replace('{}', '')`
`258`	`258`	`return not s`
	`259`	`+`
	`260`	`+# 国际站上有一种写法是这样的,相比于上面,下面的写法更加优雅(好理解)一点`
	`261`	`+class Solution:`
	`262`	`+ def isValid(self, s: str) -> bool:`
	`263`	`+ while s:`
	`264`	`+ l = len(s)`
	`265`	`+ s = s.replace('()', '').replace('[]', '').replace('{}', '')`
	`266`	`+ if l == len(s): break`
	`267`	`+ return not s`
`259`	`268`	```
`260`	`269`	`- 不断删除有效括号直到不能删除,思路简单效率低。另外,stack的方法也很简单,而且快多了。`
`261`	`270`
`@@ -264,7 +273,7 @@ class Solution:`
`264`	`273`	`def isValid(self, s: str) -> bool:`
`265`	`274`	`stack, d = [], {'{': '}', '[': ']', '(': ')'}`
`266`	`275`	`for p in s:`
`267`		`- if p in '{[(':`
	`276`	`+ if p in d:`
`268`	`277`	`stack += [p];`
`269`	`278`	`elif not (stack and d[stack.pop()] == p):`
`270`	`279`	`return False`

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