【leetcode】745. Prefix and Suffix Search

题目如下：

Given many words, words[i] has weight i.

Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix). It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.

Examples:

Input:
WordFilter(["apple"])
WordFilter.f("a", "e") // returns 0
WordFilter.f("b", "") // returns -1

Note:

1. words has length in range [1, 15000].
2. For each test case, up to words.length queries WordFilter.f may be made.
3. words[i] has length in range [1, 10].
4. prefix, suffix have lengths in range [0, 10].
5. words[i] and prefix, suffix queries consist of lowercase letters only.

解题思路：以输入apple为例，前缀和后缀分别有6个，分别为"","a","ap","app","appl","apple"，两者组合起来就是36个。words中最长的word的长度是10，那么组合就是11*11 = 121个，words的最大长度是15000，总的组合个数1815000，似乎在可以接受的范围之内。所以，只要预先把所有单词的所有前缀后缀的组合预先计算出来，那么查找的时间复杂度就是O(1)了。

代码如下：

class WordFilter(object):

    def __init__(self, words):
        """
        :type words: List[str]
        """
        self.dic = {}
        for i in range(len(words)-1,-1,-1):
            word = words[i]
            foward = ''
            reverse = word
            while len(reverse) >= 0:
                if (foward,reverse) not in self.dic:
                    self.dic[(foward,reverse)] = i
                for j in range(len(word)):
                    foward += word[j]
                    if (foward,reverse) not in self.dic:
                        self.dic[(foward,reverse)] = i
                if len(reverse) > 0:reverse = reverse[1:]
                else:break
                foward = ''

    def f(self, prefix, suffix):
        """
        :type prefix: str
        :type suffix: str
        :rtype: int
        """
        if (prefix,suffix) in self.dic:
            return self.dic[(prefix,suffix)]
        return -1