[LeetCode] 642. Design Search Autocomplete System 设计搜索自动补全系统
Design a search autocomplete system for a search engine. Users may input a sentence (at least one word and end with a special character '#'). For each character they type except '#', you need to return the top 3historical hot sentences that have prefix the same as the part of sentence already typed. Here are the specific rules:
- The hot degree for a sentence is defined as the number of times a user typed the exactly same sentence before.
- The returned top 3 hot sentences should be sorted by hot degree (The first is the hottest one). If several sentences have the same degree of hot, you need to use ASCII-code order (smaller one appears first).
- If less than 3 hot sentences exist, then just return as many as you can.
- When the input is a special character, it means the sentence ends, and in this case, you need to return an empty list.
Your job is to implement the following functions:
The constructor function:
AutocompleteSystem(String[] sentences, int[] times): This is the constructor. The input is historical data. Sentences is a string array consists of previously typed sentences. Times is the corresponding times a sentence has been typed. Your system should record these historical data.
Now, the user wants to input a new sentence. The following function will provide the next character the user types:
List<String> input(char c): The input c is the next character typed by the user. The character will only be lower-case letters ('a' to 'z'), blank space (' ') or a special character ('#'). Also, the previously typed sentence should be recorded in your system. The output will be the top 3 historical hot sentences that have prefix the same as the part of sentence already typed.
Example:
Operation: AutocompleteSystem(["i love you", "island","ironman", "i love leetcode"], [5,3,2,2])
The system have already tracked down the following sentences and their corresponding times:
"i love you" : 5 times
"island" : 3 times
"ironman" : 2 times
"i love leetcode" : 2 times
Now, the user begins another search:
Operation: input('i')
Output: ["i love you", "island","i love leetcode"]
Explanation:
There are four sentences that have prefix "i". Among them, "ironman" and "i love leetcode" have same hot degree. Since ' ' has ASCII code 32 and 'r' has ASCII code 114, "i love leetcode" should be in front of "ironman". Also we only need to output top 3 hot sentences, so "ironman" will be ignored.
Operation: input(' ')
Output: ["i love you","i love leetcode"]
Explanation:
There are only two sentences that have prefix "i ".
Operation: input('a')
Output: []
Explanation:
There are no sentences that have prefix "i a".
Operation: input('#')
Output: []
Explanation:
The user finished the input, the sentence "i a" should be saved as a historical sentence in system. And the following input will be counted as a new search.
Note:
- The input sentence will always start with a letter and end with '#', and only one blank space will exist between two words.
- The number of complete sentences that to be searched won't exceed 100. The length of each sentence including those in the historical data won't exceed 100.
- Please use double-quote instead of single-quote when you write test cases even for a character input.
- Please remember to RESET your class variables declared in class AutocompleteSystem, as static/class variables are persisted across multiple test cases. Please see here for more details.
这道题让实现一个简单的搜索自动补全系统，当我们用谷歌或者百度进行搜索时，会有这样的体验，输入些单词，搜索框会弹出一些以你输入为开头的一些完整的句子供你选择，这就是一种搜索自动补全系统。根据题目的要求，补全的句子是按之前出现的频率排列的，高频率的出现在最上面，如果频率相同，就按字母顺序来显示。输入规则是每次输入一个字符，然后返回自动补全的句子，如果遇到井字符，表示完整句子结束。那么肯定需要一个 HashMap，建立句子和其出现频率的映射，还需要一个字符串 data，用来保存之前输入过的字符。在构造函数中，给了一些句子，和其出现的次数，直接将其加入 HashMap，然后 data 初始化为空字符串。在 input 函数中，首先判读输入字符是否为井字符，如果是的话，那么表明当前的 data 字符串已经是一个完整的句子，在 HashMap 中次数加1，并且 data 清空，返回空集。否则的话将当前字符加入 data 字符串中，现在就要找出包含 data 前缀的前三高频句子了，使用优先队列来做，设计的思路是，始终用优先队列保存频率最高的三个句子，应该把频率低的或者是字母顺序大的放在队首，以便随时可以移出队列，所以应该是个最小堆，队列里放句子和其出现频率的 pair 对儿，并且根据其频率大小进行排序，要重写优先队列的 comparator。然后遍历 HashMap 中的所有句子，首先要验证当前 data 字符串是否是其前缀，没啥好的方法，就逐个字符比较，用标识符 matched，初始化为 true，如果发现不匹配，则 matched 标记为 false，并 break 掉。然后判断如果 matched 为 true 的话，说明 data 字符串是前缀，那么就把这个 pair 加入优先队列中，如果此时队列中的元素大于三个，那把队首元素移除，因为是最小堆，所以频率小的句子会被先移除。然后就是将优先队列的元素加到结果 res 中，由于先出队列的是频率小的句子，所以要加到结果 res 的末尾，参见代码如下：
class AutocompleteSystem {
public:
AutocompleteSystem(vector<string> sentences, vector<int> times) {
for (int i = 0; i < sentences.size(); ++i) {
freq[sentences[i]] += times[i];
}
data = "";
}
vector<string> input(char c) {
if (c == '#') {
++freq[data];
data = "";
return {};
}
data.push_back(c);
auto cmp = [](pair<string, int>& a, pair<string, int>& b) {
return a.second > b.second || (a.second == b.second && a.first < b.first);
};
priority_queue<pair<string, int>, vector<pair<string, int>>, decltype(cmp) > q(cmp);
for (auto f : freq) {
bool matched = true;
for (int i = 0; i < data.size(); ++i) {
if (data[i] != f.first[i]) {
matched = false;
break;
}
}
if (matched) {
q.push(f);
if (q.size() > 3) q.pop();
}
}
vector<string> res(q.size());
for (int i = q.size() - 1; i >= 0; --i) {
res[i] = q.top().first; q.pop();
}
return res;
}
private:
unordered_map<string, int> freq;
string data;
};Github 同步地址：
https://github.com/grandyang/leetcode/issues/642
类似题目：
Implement Trie (Prefix Tree)
Top K Frequent Words
参考资料：
https://leetcode.com/problems/design-search-autocomplete-system/
https://leetcode.com/problems/design-search-autocomplete-system/discuss/176550/Java-simple-solution-without-using-Trie-(only-use-HashMap-and-PriorityQueue)
https://leetcode.com/problems/design-search-autocomplete-system/discuss/105379/Straight-forward-hash-table-%2B-priority-queue-solution-in-c%2B%2B-no-trie
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