[LeetCode] 648.Replace Words 替换单词
In English, we have a concept called root, which can be followed by some other words to form another longer word - let's call this word successor. For example, the root an, followed by other, which can form another word another.
Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with the root forming it. If a successor has many roots can form it, replace it with the root with the shortest length.
You need to output the sentence after the replacement.
Example 1:
Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"
Note:
- The input will only have lower-case letters.
- 1 <= dict words number <= 1000
- 1 <= sentence words number <= 1000
- 1 <= root length <= 100
- 1 <= sentence words length <= 1000
这道题给了我们一个前缀字典,又给了一个句子,让我们将句子中较长的单词换成其前缀(如果在前缀字典中存在的话)。我们对于句子中的一个长单词如何找前缀呢,是不是可以根据第一个字母来快速定位呢,比如cattle这个单词的首字母是c,那么我们在前缀字典中找所有开头是c的前缀,为了方便查找,我们将首字母相同的前缀都放到同一个数组中,总共需要26个数组,所以我们可以定义一个二维数组来装这些前缀。还有,我们希望短前缀在长前缀的前面,因为题目中要求用最短的前缀来替换单词,所以我们可以先按单词的长度来给所有的前缀排序,然后再依次加入对应的数组中,这样就可以保证短的前缀在前面。
下面我们就要来遍历句子中的每一个单词了,由于C++中没有split函数,所以我们就采用字符串流来提取每一个单词,对于遍历到的单词,我们根据其首字母查找对应数组中所有以该首字母开始的前缀,然后直接用substr函数来提取单词中和前缀长度相同的子字符串来跟前缀比较,如果二者相等,说明可以用前缀来替换单词,然后break掉for循环。别忘了单词之前还要加上空格,参见代码如下:
解法一:
class Solution {
public:
string replaceWords(vector<string>& dict, string sentence) {
string res = "", t = "";
vector<vector<string>> v(26);
istringstream is(sentence);
sort(dict.begin(), dict.end(), [](string &a, string &b) {return a.size() < b.size();});
for (string word : dict) {
v[word[0] - 'a'].push_back(word);
}
while (is >> t) {
for (string word : v[t[0] - 'a']) {
if (t.substr(0, word.size()) == word) {
t = word;
break;
}
}
res += t + " ";
}
res.pop_back();
return res;
}
};
你以为想出了上面的解法,这道题就算做完了?? Naive! ! ! 这道题最好的解法其实是用前缀树(Trie / Prefix Tree)来做,关于前缀树使用之前有一道很好的入门题Implement Trie (Prefix Tree)。了解了前缀树的原理机制,那么我们就可以发现这道题其实很适合前缀树的特点。我们要做的就是把所有的前缀都放到前缀树里面,而且在前缀的最后一个结点的地方将标示isWord设为true,表示从根节点到当前结点是一个前缀,然后我们在遍历单词中的每一个字母,我们都在前缀树查找,如果当前字母对应的结点的表示isWord是true,我们就返回这个前缀,如果当前字母对应的结点在前缀树中不存在,我们就返回原单词,这样就能完美的解决问题了。所以啊,以后遇到了有关前缀或者类似的问题,一定不要忘了前缀树这个神器哟~
解法二:
class Solution {
public:
class TrieNode {
public:
bool isWord;
TrieNode *child[26];
TrieNode(): isWord(false) {
for (auto &a : child) a = NULL;
}
};
string replaceWords(vector<string>& dict, string sentence) {
string res = "", t = "";
istringstream is(sentence);
TrieNode *root = new TrieNode();
for (string word : dict) {
insert(root, word);
}
while (is >> t) {
if (!res.empty()) res += " ";
res += findPrefix(root, t);
}
return res;
}
void insert(TrieNode* node, string word) {
for (char c : word) {
if (!node->child[c - 'a']) node->child[c - 'a'] = new TrieNode();
node = node->child[c - 'a'];
}
node->isWord = true;
}
string findPrefix(TrieNode* node, string word) {
string cur = "";
for (char c : word) {
if (!node->child[c - 'a']) break;
cur.push_back(c);
node = node->child[c - 'a'];
if (node->isWord) return cur;
}
return word;
}
};
类似题目:
Implement Trie (Prefix Tree)
参考资料:
https://discuss.leetcode.com/topic/97203/trie-tree-concise-java-solution-easy-to-understand
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