replace Archives - Huahua's Tech Road

HTML entity parser is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.

The special characters and their entities for HTML are:

Quotation Mark: the entity is " and symbol character is ".
Single Quote Mark: the entity is ' and symbol character is '.
Ampersand: the entity is & and symbol character is &.
Greater Than Sign: the entity is > and symbol character is >.
Less Than Sign: the entity is < and symbol character is <.
Slash: the entity is &frasl; and symbol character is /.

Given the input text string to the HTML parser, you have to implement the entity parser.

Return the text after replacing the entities by the special characters.

Example 1:

Input: text = "&amp; is an HTML entity but &ambassador; is not."
Output: "& is an HTML entity but &ambassador; is not."
Explanation: The parser will replace the &amp; entity by &

Example 2:

Input: text = "and I quote: &quot;...&quot;"
Output: "and I quote: \"...\""

Example 3:

Input: text = "Stay home! Practice on Leetcode :)"
Output: "Stay home! Practice on Leetcode :)"

Example 4:

Input: text = "x &gt; y &amp;&amp; x &lt; y is always false"
Output: "x > y && x < y is always false"

Example 5:

Input: text = "leetcode.com&frasl;problemset&frasl;all"
Output: "leetcode.com/problemset/all"

Constraints:

1 <= text.length <= 10^5
The string may contain any possible characters out of all the 256 ASCII characters.

Solution: Simulation

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

string entityParser(string text) {

map<string, string> m{

{""", "\""}, {"'", "'"}, {"&", "&"},

{">", ">"}, {"<", "<"}, {"&frasl;", "/"}};

string ans;

string buf;

for (char c : text) {

buf += c;

if (buf.back() != ';') continue;

const int l = buf.size();

for (const auto& [k, v] : m) {

const int kl = k.length();

if (l >= kl && buf.substr(l - kl) == k) {

ans += buf.substr(0, l - kl) + v;

buf.clear();

break;

}

return ans + buf;

}

};

Problem

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

Solution 1: HashTable

Time complexity: O(sum(w^2))

Space complexity: O(sum(l))

// Author: Huahua, Running time: 84 ms

class Solution {

public:

string replaceWords(vector<string>& dict, string sentence) {

unordered_set<string> d(begin(dict), end(dict));

sentence.push_back(' ');

string output;

string word;

bool found = false;

for (char c : sentence) {

if (c == ' ') {

if (!output.empty()) output += ' ';

output += word;

word = "";

found = false;

continue;

}

if (found) continue;

word += c;

if (d.count(word)) {

found = true;

}

return output;

}

};

Solution2: Trie

Time complexity: O(sum(l) + n)

Space complexity: O(sum(l) * 26)

// Author: Huahua, running time: 48 ms

class TrieNode {

public:

TrieNode(): children(26), is_word(false) {}

~TrieNode() {

for (int i = 0; i < 26; ++i)

if (children[i]) {

delete children[i];

children[i] = nullptr;

}

vector<TrieNode*> children;

bool is_word;

};

class Solution {

public:

string replaceWords(vector<string>& dict, string sentence) {

TrieNode root;

for (const string& word : dict) {

TrieNode* cur = &root;

for (char c : word) {

if (!cur->children[c - 'a']) cur->children[c - 'a'] = new TrieNode();

cur = cur->children[c - 'a'];

}

cur->is_word = true;

}

string ans;

stringstream ss(sentence);

string word;

while (ss >> word) {

TrieNode* cur = &root;

bool found = false;

int len = 0;

for (char c : word) {

cur = cur->children[c - 'a'];

if (!cur) break;

++len;

if (cur->is_word) {

found = true;

break;

}

if (!ans.empty()) ans += ' ';

ans += found ? word.substr(0, len) : word;

}

return ans;

}

};

Posts tagged as “replace”

花花酱 LeetCode 1410. HTML Entity Parser

Solution: Simulation

C++

花花酱 LeetCode 648. Replace Words

Problem

Solution 1: HashTable

Solution2: Trie