问题: 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.git
方法: 对于每一个字符串和映射字符串来讲,若是映射关系中包含a->b,则map的key中包含a,则value中必包含b,反之,若是map的value中包含b,则key中必不包含a。根据如上规则,遍历整个字符串组则能够获得符合映射关系的字符串,算法复杂度为O(n)。github
具体实现:算法
class FindAndReplacePattern {
fun findAndReplacePattern(words: Array<String>, pattern: String): List<String> {
val result = mutableListOf<String>()
loop@ for (word in words) {
val map = mutableMapOf<Char, Char>()
for (i in word.indices) {
if (map.containsKey(word[i])) {
val char = map[word[i]]
if (char != pattern[i]) {
continue@loop
}
} else {
if (map.containsValue(pattern[i])) {
continue@loop
}
}
map.put(word[i], pattern[i])
}
result.add(word)
}
return result
}
}
fun main(args: Array<String>) {
val words = arrayOf("abc","deq","mee","aqq","dkd","ccc")
val pattern = "abb"
val findAndReplacePattern = FindAndReplacePattern()
CommonUtils.printArray(findAndReplacePattern.findAndReplacePattern(words, pattern).toTypedArray())
}
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有问题随时沟通bash