本文主要是介绍634 · 单词矩阵，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
链接：LintCode 炼码 - ChatGPT！更高效的学习体验！
. - 力扣（LeetCode）
题解：
class Solution {
public:
struct Trie {Trie() {next.resize(26, nullptr);end = false;}
std::vector<Trie*> next;
bool end;
};
Trie* insert_trie(vector<string>& words) {Trie* root = new (std::nothrow)Trie;for (auto& word : words) {Trie* cur = root;for (int i = 0; i < word.size(); ++i) {if (cur->next[word[i]-'a'] == nullptr) {cur->next[word[i]-'a'] = new (std::nothrow) Trie;}cur = cur->next[word[i]-'a'];}cur->end = true;}return root;
}vector<string> maxRectangle(vector<string>& words) {if (words.size() <= 0) {return {};}Trie* root = insert_trie(words);if (!root) {return {};}// 构建相同长度的表int max_word_len = 0; // std::unordered_map<int, std::unordered_set<string>> len2words;for (auto& word : words) {int len = word.size();max_word_len = max(max_word_len, len);len2words[len].insert(word);}// 回溯// 最终结果，总共字符串的长度int max_len = 0;// 记录回溯路径std::vector<string> path;// 记录最终结果std::vector<string> result;for (auto& entry : len2words) {path.clear();dfs(root, entry.first, entry.second, path, result, max_len, max_word_len);}return result;}
private:void dfs(Trie* root, int len, unordered_set<string>& words,std::vector<std::string>& path,std::vector<std::string>& result,int& max_len,int max_word_len) {// 如果当前字符的长度*宽度比最大的小，直接返回，剪枝if (len * len <= max_len) {return;}for (auto& word : words) {// 把当前字符追加到路径 中path.push_back(word);// 判断已经追加的字符串是否是正确额度auto valid = is_valid(path, root);if (valid.first) {// 获得字符串矩形的面积int tmp_len = path[0].size() * path.size();if (valid.second && tmp_len > max_len) {// 记录最大面积max_len = tmp_len;std::vector<string> tmp(path);result = std::move(tmp);}// 递归下一层dfs(root, len, words, path, result, max_len, max_word_len);}// 回溯path.pop_back();}}std::pair<int, int> is_valid(std::vector<std::string>& path, Trie* root) {int word_len = path[0].size();int row_size = path.size();bool allend = true;/*for (auto& tmp : path) {cout << tmp << " ";}cout << endl;*/// 判断字符矩阵，按照列查看是否，字符串在字典树中for (int i = 0; i < word_len; ++i) {Trie* cur = root;for (int j = 0; j < row_size; ++j) {if (cur->next[path[j][i]-'a'] == nullptr) {return std::pair<int, int>(false, false);}cur = cur->next[path[j][i]-'a'];}// 如果当前字符不是结尾if (!cur->end) {allend = false;}}//cout << " ====" << endl;return std::pair<int, int>(true , allend);}
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