Suffix tree: Difference between revisions
→{{header|D}}
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The computation time for an efficient algorithm should be <math>O(n)</math>, but such an algorithm might be difficult to implement. An easier, <math>O(n^2)</math> algorithm is accepted.
=={{header|C++}}==
{{trans|D}}
<lang cpp>#include <functional>
#include <iostream>
#include <vector>
struct Node {
std::string sub = ""; // a substring of the input string
std::vector<int> ch; // vector of child nodes
Node() {
// empty
}
Node(const std::string& sub, std::initializer_list<int> children) : sub(sub) {
ch.insert(ch.end(), children);
}
};
struct SuffixTree {
std::vector<Node> nodes;
SuffixTree(const std::string& str) {
nodes.push_back(Node{});
for (size_t i = 0; i < str.length(); i++) {
addSuffix(str.substr(i));
}
}
void visualize() {
if (nodes.size() == 0) {
std::cout << "<empty>\n";
return;
}
std::function<void(int, const std::string&)> f;
f = [&](int n, const std::string & pre) {
auto children = nodes[n].ch;
if (children.size() == 0) {
std::cout << "- " << nodes[n].sub << '\n';
return;
}
std::cout << "+ " << nodes[n].sub << '\n';
auto it = std::begin(children);
if (it != std::end(children)) do {
if (std::next(it) == std::end(children)) break;
std::cout << pre << "+-";
f(*it, pre + "| ");
it = std::next(it);
} while (true);
std::cout << pre << "+-";
f(children[children.size() - 1], pre + " ");
};
f(0, "");
}
private:
void addSuffix(const std::string & suf) {
int n = 0;
size_t i = 0;
while (i < suf.length()) {
char b = suf[i];
int x2 = 0;
int n2;
while (true) {
auto children = nodes[n].ch;
if (x2 == children.size()) {
// no matching child, remainder of suf becomes new node
n2 = nodes.size();
nodes.push_back(Node(suf.substr(i), {}));
nodes[n].ch.push_back(n2);
return;
}
n2 = children[x2];
if (nodes[n2].sub[0] == b) {
break;
}
x2++;
}
// find prefix of remaining suffix in common with child
auto sub2 = nodes[n2].sub;
size_t j = 0;
while (j < sub2.size()) {
if (suf[i + j] != sub2[j]) {
// split n2
auto n3 = n2;
// new node for the part in common
n2 = nodes.size();
nodes.push_back(Node(sub2.substr(0, j), { n3 }));
nodes[n3].sub = sub2.substr(j); // old node loses the part in common
nodes[n].ch[x2] = n2;
break; // continue down the tree
}
j++;
}
i += j; // advance past part in common
n = n2; // continue down the tree
}
}
};
int main() {
SuffixTree("banana$").visualize();
}</lang>
{{out}}
<pre>+
+-- banana$
+-+ a
| +-+ na
| | +-- na$
| | +-- $
| +-- $
+-+ na
| +-- na$
| +-- $
+-- $</pre>
=={{header|D}}==
| |||