zoukankan      html  css  js  c++  java
  • 【leetcode】1255. Maximum Score Words Formed by Letters

    题目如下:

    Given a list of words, list of  single letters (might be repeating) and score of every character.

    Return the maximum score of any valid set of words formed by using the given letters (words[i] cannot be used two or more times).

    It is not necessary to use all characters in letters and each letter can only be used once. Score of letters 'a''b''c', ... ,'z' is given by score[0]score[1], ... , score[25] respectively.

    Example 1:

    Input: words = ["dog","cat","dad","good"], letters = ["a","a","c","d","d","d","g","o","o"], 
    score = [1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0] Output: 23 Explanation: Score a=1, c=9, d=5, g=3, o=2 Given letters, we can form the words "dad" (5+1+5) and "good" (3+2+2+5) with a score of 23. Words "dad" and "dog" only get a score of 21.

    Example 2:

    Input: words = ["xxxz","ax","bx","cx"], letters = ["z","a","b","c","x","x","x"], 
    score = [4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10] Output: 27 Explanation: Score a=4, b=4, c=4, x=5, z=10 Given letters, we can form the words "ax" (4+5), "bx" (4+5) and "cx" (4+5) with a score of 27. Word "xxxz" only get a score of 25.

    Example 3:

    Input: words = ["leetcode"], letters = ["l","e","t","c","o","d"], score = [0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0]
    Output: 0
    Explanation:
    Letter "e" can only be used once. 

    Constraints:

    • 1 <= words.length <= 14
    • 1 <= words[i].length <= 15
    • 1 <= letters.length <= 100
    • letters[i].length == 1
    • score.length == 26
    • 0 <= score[i] <= 10
    • words[i]letters[i] contains only lower case English letters.

    解题思路:这题也能算hard级别?words.length 最大才14,那么总共有2^14次方种组合,全列举出来求最大值即可。

    代码如下:

    class Solution(object):
        def maxScoreWords(self, words, letters, score):
            """
            :type words: List[str]
            :type letters: List[str]
            :type score: List[int]
            :rtype: int
            """
            def checkValid(string,dic):
                for i in string:
                    if i not in dic or string.count(i) > dic[i]:
                        return False
                return True
            def calc(string,dic):
                uniq = set(list(string))
                count = 0
                for i in uniq:
                    count += string.count(i) * score[ord(i) - ord('a')]
                return count
    
            dic = {}
            for i in letters:
                dic[i] = dic.setdefault(i,0) + 1
            queue = []
            for i in range(len(words)):
                if checkValid(words[i],dic):
                    queue.append((i,words[i]))
            res = 0
            while len(queue) > 0:
                inx,word = queue.pop(0)
                res = max(res,calc(word,dic))
                for i in range(inx+1,len(words)):
                    if checkValid(word + words[i],dic):
                        queue.append((i,word + words[i]))
            return res
  • 相关阅读:
    Sort
    RAID
    LeetCode总结 -- 一维动态规划篇
    Count and Say
    Dynamic Programming Introduction
    Mongodb与Redis应用指标对比
    精通有状态vs无状态(Stateful vs Stateless)—Immutable模式之姐妹篇
    Windows 安装 pytorch3d
    GitHub 图片无法显示 或 gist 无法访问
    LaTeX符号表,数学公式速查必备
  • 原文地址:https://www.cnblogs.com/seyjs/p/11841269.html
Copyright © 2011-2022 走看看