748. Shortest Completing Word

最新推荐文章于 2022-12-09 14:36:04 发布

转载最新推荐文章于 2022-12-09 14:36:04 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/bernieloveslife/p/9835877.html

本文介绍了一种算法，用于从给定字典中找出最短的单词，该单词包含特定车牌字符串中的所有字母，考虑大小写不敏感。通过两个解决方案的对比，展示了如何优化排序过程，以提高算法效率。

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Find the minimum length word from a given dictionary words, which has all the letters from the string licensePlate. Such a word is said to complete the given string licensePlate

Here, for letters we ignore case. For example, "P" on the licensePlate still matches "p" on the word.

It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.

The license plate might have the same letter occurring multiple times. For example, given a licensePlate of "PP", the word "pair" does not complete the licensePlate, but the word "supper" does.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
Output: "steps"
Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
Note that the answer is not "step", because the letter "s" must occur in the word twice.
Also note that we ignored case for the purposes of comparing whether a letter exists in the word.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
Output: "pest"
Explanation: There are 3 smallest length words that contains the letters "s".
We return the one that occurred first.

Note:

licensePlate will be a string with length in range [1, 7].
licensePlate will contain digits, spaces, or letters (uppercase or lowercase).
words will have a length in the range [10, 1000].
Every words[i] will consist of lowercase letters, and have length in range [1, 15].

Solution1:(TLE)

class Solution:
    def shortestCompletingWord(self, licensePlate, words):
        """
        :type licensePlate: str
        :type words: List[str]
        :rtype: str
        """
        dic = []
        for i in licensePlate:
            if i.isalpha():
                dic.append(i.lower())
        # print(dic)
        def bubble_sort(l):
            for i in range(len(l)-1):
                for j in range(len(l)-1-i):
                    if len(l[j])>len(l[j+1]):
                        l[j],l[j+1] = l[j+1],l[j]
                # print(l)
        def judge(a):
            # print('a',a)
            d = {}
            for i in a:
                if i not in d:
                    d[i] = 1
                else:
                    d[i] += 1
            # print('d',d)
            for i in dic:
                if i not in d:
                    return False
                if d[i]<1:
                    return False
                d[i] -= 1
            return True
        bubble_sort(words)
        # print(words)
        for word in words:
            if judge(word):
                return word

Solution2:

class Solution:
    def shortestCompletingWord(self, licensePlate, words):
        """
        :type licensePlate: str
        :type words: List[str]
        :rtype: str
        """
        dic = []
        for i in licensePlate:
            if i.isalpha():
                dic.append(i.lower())
        # print(dic)
        def judge(a):
            # print('a',a)
            d = {}
            for i in a:
                if i not in d:
                    d[i] = 1
                else:
                    d[i] += 1
            # print('d',d)
            for i in dic:
                if i not in d:
                    return False
                if d[i]<1:
                    return False
                d[i] -= 1
            return True
        words.sort(key = lambda x:len(x))
        # print(words)
        for word in words:
            if judge(word):
                return word

只是成了sort就过了，注意sort本身就是稳定的排序方法了。

转载于:https://www.cnblogs.com/bernieloveslife/p/9835877.html