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描述
Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order. A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word. Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
Example 1:
Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
Example 2:
Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]
Note:
1 <= pattern.length <= 20
1 <= words.length <= 50
words[i].length == pattern.length
pattern and words[i] are lowercase English letters.
解析
根据题意，给定一个字符串列表 words 和一个字符串 pattern ，返回一个匹配 pattern 的 words[i] 列表。如果存在字母排列 p 的，将模式中的每个字母 x 替换为 p(x) 之后，我们就得到了所需的单词。从字母到字母的映射：每个字母都映射到另一个字母，没有两个字母映射到同一个字母。
其实这道题很简单，就是遍历所有的 words 中的单词，然后比较单词和 pattern 是不是能够匹配到 pattern ，而这个关键就是字母之间的映射，只要满足了从单词到 pattern 的字符映射我们就认为是能够匹配的，否则说明不能够匹配。
时间复杂度为 O(N) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def findAndReplacePattern(self, words, pattern):
"""
:type words: List[str]
:type pattern: str
:rtype: List[str]
"""
result = []
for word in words:
d = collections.defaultdict()
N = len(word)
i = 0
while i < N:
c,t = word[i],pattern[i]
if c not in d:
if t in d.values():
break
else:
d[c] = t
elif d[c] != t:
break
i += 1
if i == N:
result.append(word)
return result
运行结果
Runtime: 23 ms, faster than 82.11% of Python online submissions for Find and Replace Pattern.
Memory Usage: 13.8 MB, less than 11.38% of Python online submissions for Find and Replace Pattern.
原题链接
https://leetcode.com/problems/find-and-replace-pattern/
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