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].

思路:(看讨论的,叫做直方图解法,特别巧妙)

class Solution {
    inline bool metLicense(vector<int> vec) {  //判断vec中是否存在>0的元素
        for (auto i : vec) {
            if (i > 0) 
                return false;  //细节:注意一定要>0, !=0都不行.
        }                      //因为题目要求的是plate里面出现的一定要满足,但不限制出现plate以外的字符.
        return true;
    }
public:
    string shortestCompletingWord(string licensePlate, vector<string>& words) {
        string res;
        vector<int> dict(26);
        for (auto c : licensePlate) { //将plate出现的字符映射到dict数组的0~25号位置,
            if (isalpha(c)) //若是字符
                dict[tolower(c) - 'a']++;  //全部转为小写,检测到一个,相应位置就+1
        }
        for (auto s : words) {  //遍历每个string
            vector<int> tmp = dict; //tmp复制dict,用于检测
            for (auto c : s) {  //对string的每个字符
                if (isalpha(c)) tmp[tolower(c) - 'a']--;  //tmp相应位置-1
            }
            //借助辅助函数metLicense判断是否满足要求
            //此外若发现更短的串,或者是初始串的时候
            if (metLicense(tmp) && (s.length() < res.length() || res.length() == 0))
                res = s;
        }
        return res;
    }
};