在这个问题中，我们得到了一个输入字符串和一个数组arr []。我们的任务是查找字符串中数组所有单词的所有匹配项。为此，我们将使用Aho-Corasick算法进行模式搜索。
字符串和模式搜索在编程中很重要。而且在编程中，算法越好，它可以具有更多的实际用途。Aho-Corasick算法是一种非常重要且功能强大的算法，可简化字符串搜索。这是一种字典匹配算法，可以同时匹配所有字符串。该算法将Trie数据结构用于其实现。
Trie是一种前缀树或数字搜索树，其中每个边缘都用某个字母标记（每个输出边缘具有不同的字母）。
让我们以一个例子来了解Aho-Corasick算法
输入值
string = "bheythisghisanexample"
arr[] = {"hey", "this", "is", "an", “example”}输出结果
Word hey starts from 2 Word this starts from 5 Word is starts from 11 Word an starts from 13 Word example starts from 15
该算法的时间复杂度为O（N + L + Z），其中N =字符串/文本输入的长度
L =关键字的长度（数组中的单词）
Z =比赛次数。
可以通过以下简单步骤构建Aho-Corasick算法
使用队列构造特里，以便我们可以将队列中的每个字符作为节点“特里”弹出。
将故障链接（后缀链接）构造为可以存储下一个和当前字符的数组
将输出链接构造为数组以存储匹配的单词
建立一个遍历函数(FindNextState)以检查所有字符。
失败链接（后缀链接） -当我们击中无法继续读取字符的字符串部分时，我们通过跟随后缀链接来回退，以尝试尽可能多地保留上下文。简而言之，当当前角色在Trie中没有边时，它将存储所有跟随的边。
输出链接-它始终指向与当前状态中存在的最长单词相对应的节点，我们确保使用输出链接将所有模式链接在一起。
#include<iostream>
#include <string.h>
#include<algorithm>
#include<queue>
using namespace std;
const int MaxStates = 6 * 50 + 10;
const int MaxChars = 26;
int OccurenceOfWords[MaxStates];
int FF[MaxStates];
int GotoFunction[MaxStates][MaxChars];
int BuildMatchingMachine(const vector<string> &words, char lowestChar = 'a', char highestChar = 'z'){
memset(OccurenceOfWords, 0, sizeof OccurenceOfWords);
memset(FF, -1, sizeof FF);
memset(GotoFunction, -1, sizeof GotoFunction);
int states = 1;
for (int i = 0; i < words.size(); ++i){
const string &keyword = words[i];
int currentState = 0;
for (int j = 0; j < keyword.size(); ++j){
int c = keyword[j] - lowestChar;
if (GotoFunction[currentState][c] == -1){
GotoFunction[currentState][c] = states++;
}
currentState = GotoFunction[currentState][c];
}
OccurenceOfWords[currentState] |= (1 << i);
}
for (int c = 0; c < MaxChars; ++c){
if (GotoFunction[0][c] == -1){
GotoFunction[0][c] = 0;
}
}
queue<int> q;
for (int c = 0; c <= highestChar - lowestChar; ++c){
if (GotoFunction[0][c] != -1 && GotoFunction[0][c] != 0){
FF[GotoFunction[0][c]] = 0;
q.push(GotoFunction[0][c]);
}
}
while (q.size()){
int state = q.front();
q.pop();
for (int c = 0; c <= highestChar - lowestChar; ++c){
if (GotoFunction[state][c] != -1){
int failure = FF[state];
while (GotoFunction[failure][c] == -1){
failure = FF[failure];
}
failure = GotoFunction[failure][c];
FF[GotoFunction[state][c]] = failure;
OccurenceOfWords[GotoFunction[state][c]] |= OccurenceOfWords[failure];
q.push(GotoFunction[state][c]);
}
}
}
return states;
}
int FindNextState(int currentState, char nextInput, char lowestChar = 'a'){
int answer = currentState;
int c = nextInput - lowestChar;
while (GotoFunction[answer][c] == -1){
answer = FF[answer];
}
return GotoFunction[answer][c];
}
vector<int> FindWordCount(string str, vector<string> keywords, char lowestChar = 'a', char highestChar = 'z') {
BuildMatchingMachine(keywords, lowestChar, highestChar);
int currentState = 0;
vector<int> retVal;
for (int i = 0; i < str.size(); ++i){
currentState = FindNextState(currentState, str[i], lowestChar);
if (OccurenceOfWords[currentState] == 0)
continue;
for (int j = 0; j < keywords.size(); ++j){
if (OccurenceOfWords[currentState] & (1 << j)){
retVal.insert(retVal.begin(), i - keywords[j].size() + 1);
}
}
}
return retVal;
}
int main(){
vector<string> keywords;
keywords.push_back("All");
keywords.push_back("she");
keywords.push_back("is");
string str = "Allisheall";
cout<<"The occurrences of all words in the string ' "<<str<<" ' are \n";
vector<int> states = FindWordCount(str, keywords);
for(int i=0; i < keywords.size(); i++){
cout<<"Word "<<keywords.at(i)<<' ';
cout<<"starts at "<<states.at(i)+1<<' ';
cout<<"And ends at "<<states.at(i)+keywords.at(i).size()+1<<endl;
}
}输出结果
The occurrences of all words in the string ' Allisheall ' are Word All starts at 5 And ends at 8 Word she starts at 4 And ends at 7 Word is starts at 1 And ends at 3