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  • 【leetcode】1178. Number of Valid Words for Each Puzzle

    题目如下:

    With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:
    • word contains the first letter of puzzle.
    • For each letter in word, that letter is in puzzle.
      For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).
    Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

    Example :

    Input: 
    words = ["aaaa","asas","able","ability","actt","actor","access"], 
    puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
    Output: [1,1,3,2,4,0]
    Explanation:
    1 valid word for "aboveyz" : "aaaa" 
    1 valid word for "abrodyz" : "aaaa"
    3 valid words for "abslute" : "aaaa", "asas", "able"
    2 valid words for "absoryz" : "aaaa", "asas"
    4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
    There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.
    

    Constraints:

    • 1 <= words.length <= 10^5
    • 4 <= words[i].length <= 50
    • 1 <= puzzles.length <= 10^4
    • puzzles[i].length == 7
    • words[i][j]puzzles[i][j] are English lowercase letters.
    • Each puzzles[i] doesn't contain repeated characters.

    解题思路:本题可以采用逆向思维法。首先把words中的每个单词都做去掉重复字符处理,然后按字典序重新把字符排序,得到新的单词,例如 ealebb 经过处理后就是 able。接下来对puzzles中的每个单词除第一个字符外的子字符串做同样的处理,例如 gawbxyz 处理后的结果是 g + abwxyz 两部分。对于满足gawbxyz谜底的单词要具备的条件是 第一个字符g是必定存在,同时后面的abwxyz 任意选择0~6个字符,这要判断这些组合有哪些在处理words中出现过,即可得puzzles中每个puzzle的结果。因为puzzles[i].length == 7,所以最多只会有2^6次方中组合,不会存在性能问题。

    代码如下:

    class Solution(object):
        def findNumOfValidWords(self, words, puzzles):
            """
            :type words: List[str]
            :type puzzles: List[str]
            :rtype: List[int]
            """
            dic_words = {}
            for word in words:
                newword = ''.join(sorted(list(set(list(word)))))
                dic_words[newword] = dic_words.setdefault(newword,0) + 1
            res = []
            for puzzle in puzzles:
                count = 0
                first = puzzle[0]
                puzzle = ''.join(sorted(list(set(list(puzzle[1:])))))
                if puzzle.find(first) != -1:
                    inx = puzzle.find(first)
                    puzzle = puzzle[:inx] + puzzle[inx+1:]
                if first in dic_words:
                    count += dic_words[first]
                queue = []
                for (i,v) in enumerate(puzzle):
                    queue.append((v,i))
                while len(queue) > 0:
                    v,i = queue.pop(0)
                    if i == len(puzzle) - 1:
                        key = ''.join(sorted(list(first + v)))
                        if key in dic_words:
                            count += dic_words[key]
                        continue
                    queue.append((v,i+1))
                    queue.append((v + puzzle[i+1],i+1))
                res.append(count)
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/11475415.html
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