LeetCode周赛#104 Q3 Word Subsets (map)

题目来源：https://leetcode.com/contest/weekly-contest-104/problems/word-subsets/

问题描述

916. Word Subsets

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 B, b 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].

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题意

给定字符串集合A和集合B，定义字符串b是字符串a的”subset”当且仅当b中字母出现次数小于等于a中对应字母出现的次数。对于A中的每个字符串，判断B中所有字符串是不是都是它的”subset”.

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思路

这次周赛的题目好多都是用map做的。用map记录B中各个字符串中每个字符出现的次数，再用一个map求这些map的并。然后再用map记录A中各个字符串每个字符出现的次数，与前面求并的map作比较即可。

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代码

class Solution {
public:
    map<char, int> mp;
    
    void init(vector<string> &B)
    {
        int i, j, len, lB = B.size();
        map<char, int> bp;
        map<char, int>::iterator it;
        string b;
        for (i=0; i<lB; i++)
        {
            b = B.at(i);
            len = b.size();
            bp.clear();
            for (j=0; j<len; j++)
            {
                bp[b[j]]++;
            }
            for (it=bp.begin(); it!=bp.end(); it++)
            {
                if (mp[it->first] < it->second)
                {
                    mp[it->first] = it->second;
                }
            }
        }
    }
    
    vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
        init(B);
        int i, j, len, lA = A.size(), lB = B.size();
        string a;
        map<char, int> ap, bp;
        map<char, int>::iterator it;
        bool is_ans = true;
        vector<string> ans;
        for (i=0; i<lA; i++)
        {
            is_ans = true;
            a = A[i];
            len = a.size();
            ap.clear();
            for (j=0; j<len; j++)
            {
                ap[a[j]]++;
            }
            for (it=mp.begin(); it!=mp.end(); it++)
            {
                if (ap[it->first] < it->second)
                {
                    is_ans = false;
                    break;
                }
            }
            if (is_ans)
            {
                ans.push_back(a);
            }
        }
        return ans;
    }
};