748. Shortest Completing Word

最新推荐文章于 2022-03-30 21:26:34 发布

Ulricalin

最新推荐文章于 2022-03-30 21:26:34 发布

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分类专栏： algorithm-design

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本文介绍了一种算法，用于从给定的字典中找到最短的单词，该单词包含指定字符串中的所有字母，考虑字母大小写不敏感的情况。通过哈希表记录目标字符串中每个字母的出现次数，并对比字典中的单词来确定是否包含所有必需的字母。

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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].

Hint:
要找包含给出模板所有字母的最短单词，可以使用哈希表记录模板中每个字母的个数，再对应每个单词中每个字母出现的个数，如果后者中每个字母出现个数大于或等于前者，则改单词是完全包含模板的，找出其中长度最短的即可。

class Solution {
public:
    string shortestCompletingWord(string licensePlate, vector<string>& words) {
        licensePlate = toLower(licensePlate);
        int mymap[26];
        for (int i = 0; i < 26; i++) {
            mymap[i] = 0;
        }
        for (int i = 0; i < licensePlate.length(); i++) {
            if (licensePlate[i] <= 'z' && licensePlate[i] >= 'a') {
                mymap[licensePlate[i]-'a']++;
            }
        }
        string result;
        int minlength = 1001;
        for (auto word : words) {
            if (matched(word, mymap) && word.length() < minlength) {
                minlength = word.length();
                result = word;
            }
        }
        return result;
    }
    bool matched(string word, int* mymap) {
        int tmap[26];
        for (int i = 0; i < 26; i++) {
            tmap[i] = mymap[i];
        }
        for (int i = 0; i < word.length(); i++) {
            tmap[word[i]-'a']--;
        }
        for (int i = 0; i < 26; i++) {
            if (tmap[i] > 0 ) {
                return false;
            }
        }
        return true;
    }
    string toLower(string s) {
        for (int i = 0; i < s.length(); i++) {
            if (s[i] >= 'A' && s[i] <= 'Z') {
                s[i] = s[i]-'A'+'a';
            }
        }
        return s;
    }
};