英文原文
Given a pattern
and a string s
, find if s
follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern
and a non-empty word in s
.
Example 1:
Input: pattern = "abba", s = "dog cat cat dog" Output: true
Example 2:
Input: pattern = "abba", s = "dog cat cat fish" Output: false
Example 3:
Input: pattern = "aaaa", s = "dog cat cat dog" Output: false
Example 4:
Input: pattern = "abba", s = "dog dog dog dog" Output: false
Constraints:
1 <= pattern.length <= 300
pattern
contains only lower-case English letters.1 <= s.length <= 3000
s
contains only lower-case English letters and spaces' '
.s
does not contain any leading or trailing spaces.- All the words in
s
are separated by a single space.
中文题目
给定一种规律 pattern
和一个字符串 str
,判断 str
是否遵循相同的规律。
这里的 遵循 指完全匹配,例如, pattern
里的每个字母和字符串 str
中的每个非空单词之间存在着双向连接的对应规律。
示例1:
输入: pattern ="abba"
, str ="dog cat cat dog"
输出: true
示例 2:
输入:pattern ="abba"
, str ="dog cat cat fish"
输出: false
示例 3:
输入: pattern ="aaaa"
, str ="dog cat cat dog"
输出: false
示例 4:
输入: pattern ="abba"
, str ="dog dog dog dog"
输出: false
说明:
你可以假设 pattern
只包含小写字母, str
包含了由单个空格分隔的小写字母。
通过代码
高赞题解
class Solution:
def wordPattern(self, pattern: str, str: str) -> bool:
res=str.split()
return list(map(pattern.index, pattern))==list(map(res.index,res))
统计信息
通过次数 | 提交次数 | AC比率 |
---|---|---|
87506 | 192101 | 45.6% |
提交历史
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题目 | 难度 |
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同构字符串 | 简单 |
单词规律 II | 中等 |