2018年9月16日
#890. Find and Replace Pattern
问题描述:
You have a list of words
and a pattern
, and you want to know which words in words
matches the pattern.
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.)
Return a list of the words in words
that match the given pattern.
You may return the answer in any order.
样例:
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.
问题分析:
本题难度为Medium!已给出的函数定义为:
class Solution:
def findAndReplacePattern(self, words, pattern):
"""
:type words: List[str]
:type pattern: str
:rtype: List[str]
"""
其中words是字符串数组,pattern为字符串;返回值是一个字符串数组。
算法如下:
遍历words的每一个字符串元素与pattern进行匹配,若符合,则存储该字符串到数组中;
进行匹配判断时,创建两个字典m1、m2,分别存储words字符串元素的字符到pattern字符串的字符的映射和pattern字符串的字符到words字符串元素的字符的映射,若关键字key不存在,则添加映射;
判断(m1[w],m2[p])==(p,w)是否成立,即映射是否为一一映射,若不是,则不匹配,若全部成立,则匹配;
将匹配的字符串添加到数组中;
代码实现如下:
class Solution:
def findAndReplacePattern(self, words, pattern):
"""
:type words: List[str]
:type pattern: str
:rtype: List[str]
"""
res=[]
for word in words:
m1, m2 = {}, {}
isMatch=True
for i in range(len(pattern)):
w = word[i]
p = pattern[i]
if w not in m1:
m1[w] = p
if p not in m2:
m2[p] = w
if (m1[w], m2[p]) != (p, w):
isMatch=False
break
if isMatch:
res.append(word)
return res
由于python为我们提供了两个便捷强大的函数zip()和filter(),因此代码可以进一步简化,以下代码是官方提供的标准代码:
class Solution:
def findAndReplacePattern(self, words, pattern):
"""
:type words: List[str]
:type pattern: str
:rtype: List[str]
"""
def match(word):
m1, m2 = {}, {}
for w, p in list(zip(word, pattern)):
if w not in m1:
m1[w] = p
if p not in m2:
m2[p] = w
if (m1[w], m2[p]) != (p, w):
return False
return True
return list(filter(match, words))