【Leetcode】745. Prefix and Suffix Search 745. 前缀和后缀搜索

解法

感觉好像单纯是翻译一下官解哈哈哈哈

解法一：成对的TRIE

对于长度相同的前后缀，比如abc和xyz，可以表示为('a','z')、('b','y')、('c','x')
但是对于长度不等的，可以用None来表示空位
综上就是把TRIE的搜索路径表示为itertools.izip_longest(prefix,suffix[::-1])
难点在于TRIE的构造，见注释：

class WordFilter(object):

    def __init__(self, words):
        """
        :type words: List[str]
        """
        from collections import defaultdict
        Trie = lambda:defaultdict(Trie)
        self.trie = Trie()
        WEIGHT = 0
        for weight,word in enumerate(words):
            cur = self.trie
            cur[WEIGHT] = weight
            for i,w in enumerate(word):
                # 现在已经遍历了深度i，（即prefix和suffix里短的那个长为i），要往i+1迈进
                
                # 那些`suffix`还维持在长度i，但是`prefix`更长的情况
                tmp = cur
                for c in word[i:]:
                    tmp = tmp[c,None]
                    tmp[WEIGHT] = weight
                
                # 那些`prefix`还维持在长度i，但是`suffix`更长的情况
                tmp = cur
                for c in word[:-i or None][::-1]:
                    tmp = tmp[None,c]
                    tmp[WEIGHT] = weight
                
                # 更新长度相等的时候的值
                cur = cur[w,word[-i-1]]
                cur[WEIGHT] = weight
                
    def f(self, prefix, suffix):
        """
        :type prefix: str
        :type suffix: str
        :rtype: int
        """
        root = reduce(dict.__getitem__, itertools.izip_longest(prefix,suffix[::-1]), self.trie)
        if 0 in root:
            return root[0]
        return -1

解法二：拼接后缀

对于单词apple，的可能后缀e、ple等，可以组成新的字符串e#apple、ple#apple
把这两个字符串加到TRIE里，那么搜索的时候就是直接搜suffix#prefix在不在TRIE里就行了
比如prefix = app, suffix = ple时，搜ple#app就行搜到了

class WordFilter(object):

    def __init__(self, words):
        """
        :type words: List[str]
        """
        from collections import defaultdict
        import functools
        Trie = lambda:defaultdict(Trie)
        self.trie = Trie()
        WEIGHT = 0
        def deal(x,y,weight):
            x[WEIGHT] = weight
            return x[y]
        
        for weight,word in enumerate(words):
            cur = self.trie
            cur[WEIGHT] = weight
            for i in range(len(word)+1):
                root = reduce(functools.partial(deal,weight=weight),word[i:]+'#'+word,self.trie)
                root[WEIGHT] = weight
                
    def f(self, prefix, suffix):
        """
        :type prefix: str
        :type suffix: str
        :rtype: int
        """
        root = reduce(dict.__getitem__, suffix+"#"+prefix, self.trie)
        if 0 in root:
            return root[0]
        return -1