Leetcode-916. Word Subsets-(Medium)

一、问题描述

　　

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].

 

　　解释：

　　给定字符串数组B，如果B中的每个元素串中的每个字符都在一个字符串中（计算重复）；那么说明这个字符串符合条件

　　输入两个字符串数组，A、B；B为模式字符串数组；判断A中的多少个字符串符合预期？

 

二、解答

　　

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#include <cstdlib> #include <iostream> #include <string> #include <vector> #include <sstream> #include <algorithm> #include <set> #include <map> using namespace std;     bool isUniversal(string& word, std::map<char, int> chr_map) {     for(auto c:word){         if(chr_map.find(c) != chr_map.end())             chr_map[c] -= 1;     }           std::map<char, int>::iterator it = chr_map.begin();     while (it != chr_map.end()) {         if((it->second) > 0)             return false;         it++;     }           return true; }   class Solution { public:     vector<string> wordSubsets(vector<string>& A, vector<string>& B) {         std::map<char, int> chr_map;         std::vector<string> result;         for(auto word:B){             std::map<char, int> chr_temp_map;             for(auto c:word){                 chr_temp_map[c] += 1;             }             for (auto& kv : chr_temp_map) {                 chr_map[kv.first] = std::max(kv.second, chr_map[kv.first]);             }         }                   for(auto w: A){             if(isUniversal(w, chr_map))             {                 result.push_back(w);             }         }         return result;     } };       void test() {     std::map<char, int> chr_map;     std::vector<string> result{"abc","bcd","cde"};     for(auto word:result){         for(auto c:word){             chr_map[c] += 1;             cout<<chr_map[c]<<endl;         }     }           std::map<char, int> chr_map_copy(chr_map);       }     int main(int argc, const char * argv[]) {     // insert code here... //    test();     Solution s;     vector<string> A{"amazon","apple","facebook","google","leetcode"};     vector<string> B{"lo","eo"};     s.wordSubsets(A, B);           return 0; }

　　

三、总结

　　这道题目没有特殊，主要是理解题意。

　　用到std::<string> map，for 遍历 c++11的语法

　　编译需要添加 -std=c++11 参数