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给你一个数组 favoriteCompanies ,其中 favoriteCompanies[i] 是第 i 名用户收藏的公司清单(下标从 0 开始)。
请找出不是其他任何人收藏的公司清单的子集的收藏清单,并返回该清单下标。下标需要按升序排列。
示例 1:输入:favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]输出:[0,1,4] 解释:favoriteCompanies[2]=["google","facebook"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 的子集。favoriteCompanies[3]=["google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 和 favoriteCompanies[1]=["google","microsoft"] 的子集。其余的收藏清单均不是其他任何人收藏的公司清单的子集,因此,答案为 [0,1,4] 。示例 2:输入:favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]输出:[0,1] 解释:favoriteCompanies[2]=["facebook","google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 的子集,因此,答案为 [0,1] 。示例 3:输入:favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]输出:[0,1,2,3] 提示:1 <= favoriteCompanies.length <= 1001 <= favoriteCompanies[i].length <= 5001 <= favoriteCompanies[i][j].length <= 20favoriteCompanies[i] 中的所有字符串 各不相同 。用户收藏的公司清单也 各不相同 ,也就是说,即便我们按字母顺序排序每个清单,favoriteCompanies[i] != favoriteCompanies[j] 仍然成立。所有字符串仅包含小写英文字母。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
template <class InputIterator1, class InputIterator2>
bool includes ( InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2 );
template <class InputIterator1, class InputIterator2, class Compare>
bool includes ( InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, Compare comp );
Test whether sorted range includes another sorted range
Returns true if the sorted range [first1,last1) contains all the elements in the sorted range [first2,last2).
includes
准备includes
判断class Solution { public: vector peopleIndexes(vector>& favoriteCompanies) { map >> m;//len,排序后的字符 for(auto& fc : favoriteCompanies) { sort(fc.begin(), fc.end()); m[int(fc.size())].insert(fc); } vector ans; int len; bool flag; for(int i = 0; i < favoriteCompanies.size(); ++i) { len = favoriteCompanies[i].size(); //单词个数,只需要在大于它的集合里查找 auto it = m.lower_bound(len+1); flag = true; for(auto iter = it; iter != m.end(); ++iter) { for(auto sit = iter->second.begin(); sit != iter->second.end(); ++sit) { if(includes(sit->begin(),sit->end(),favoriteCompanies[i].begin(),favoriteCompanies[i].end())) { flag = false; break; } } if(!flag) break; } if(flag) ans.push_back(i); } return ans; }};
464 ms 52.3 MB