• [LeetCode] 916. Word Subsets 单词子集合

  • We are given two arrays `A` and `B` of words.  Each word is a string of lowercase letters.

    Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity.  For example, "wrr" is a subset of "warrior", but is not a subset of "world".

    Now say a word a from A is universal if for every b in Bb is a subset of a.

    Return a list of all universal words in A.  You can return the words in any order.

    Example 1:

    Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
    Output: ["facebook","google","leetcode"]
    

    Example 2:

    Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
    Output: ["apple","google","leetcode"]
    

    Example 3:

    Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
    Output: ["facebook","google"]
    

    Example 4:

    Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
    Output: ["google","leetcode"]
    

    Example 5:

    Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
    Output: ["facebook","leetcode"]
    

    Note:

    1. 1 <= A.length, B.length <= 10000
    2. 1 <= A[i].length, B[i].length <= 10
    3. A[i] and B[i] consist only of lowercase letters.
    4. All words in A[i] are unique: there isn't i != j with A[i] == A[j].

    这道题定义了两个单词之间的一种子集合关系,就是说假如单词b中的每个字母都在单词a中出现了(包括重复字母),就说单词b是单词a的子集合。现在给了两个单词集合A和B,让找出集合A中的所有满足要求的单词,使得集合B中的所有单词都是其子集合。配合上题目中给的一堆例子,意思并不难理解,根据子集合的定义关系,其实就是说若单词a中的每个字母的出现次数都大于等于单词b中每个字母的出现次数,单词b就一定是a的子集合。现在由于集合B中的所有单词都必须是A中某个单词的子集合,那么其实只要对于每个字母,都统计出集合B中某个单词中出现的最大次数,比如对于这个例子,B=["eo","oo"],其中e最多出现1次,而o最多出现2次,那么只要集合A中有单词的e出现不少1次,o出现不少于2次,则集合B中的所有单词一定都是其子集合。这就是本题的解题思路,这里使用一个大小为 26 的一维数组 charCnt 来统计集合B中每个字母的最大出现次数,而将统计每个单词的字母次数的操作放到一个子函数 helper 中,当 charCnt 数组更新完毕后,下面就开始检验集合A中的所有单词了。对于每个遍历到的单词,还是要先统计其每个字母的出现次数,然后跟 charCnt 中每个位置上的数字比较,只要均大于等于 charCnt 中的数字,就可以加入到结果 res 中了,参见代码如下:
    class Solution {
    public:
        vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
            vector<string> res;
            vector<int> charCnt(26);
            for (string &b : B) {
                vector<int> t = helper(b);
                for (int i = 0; i < 26; ++i) {
                    charCnt[i] = max(charCnt[i], t[i]);
                }
            }
            for (string &a : A) {
                vector<int> t = helper(a);
                int i = 0;
                for (; i < 26; ++i) {
                    if (t[i] < charCnt[i]) break;
                }
                if (i == 26) res.push_back(a);
            }
            return res;
        }
        vector<int> helper(string& word) {
            vector<int> res(26);
            for (char c : word) ++res[c - 'a'];
            return res;
        }
    };
    

    Github 同步地址:

    https://github.com/grandyang/leetcode/issues/916


    参考资料:

    https://leetcode.com/problems/word-subsets/

    https://leetcode.com/problems/word-subsets/discuss/175854/C%2B%2BJavaPython-Straight-Forward


    [LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)
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  • 原文地址:https://www.cnblogs.com/grandyang/p/11623684.html
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