这是另一种字符串匹配方法。在这种方法中,我们正在使用向量搜索子字符串。
在C ++中,我们可以使用标准库轻松创建矢量。我们将主字符串和要搜索的字符串作为向量,然后将其搜索到主字符串中。找到一个匹配项后,该函数将返回地址,并将其从主字符串中删除。因此,在下一次迭代中,它从位置0开始并再次搜索。
对于多次出现,我们使用循环并重复搜索匹配项,然后返回位置。
Input: Main String: “ABAAABCDBBABCDDEBCABC”, Pattern “ABC” Output: Pattern found at position: 4 Pattern found at position: 10 Pattern found at position: 18
输入 -主要文本和子字符串。
输出-找到样式的位置
Begin p := starting point of the main string while r is not at the end of substr and p is not at the end of main, do r := starting of substr while item at position p & r are not same, and p in main, do p := p + 1 i := i + 1 done q := p while item at pos p & r are same, and r in substr and p in main, do p := p + 1 i := i + 1 r := r + 1 done if r exceeds the substr, then delete first occurrence of substr from main return the position where substr is found if p exceeds main string, then return 0 q := q + 1 p := q done End
#include <iostream>
#include <string>
#include <vector>
using namespace std;
void take_string(vector<char> &string){
char c;
while(true){
c = getchar();
if(c == '\n'){
break;
}
string.push_back(c);
}
}
void display(vector<char> string){
for(int i = 0; i<string.size(); i++){
cout << string[i];
}
}
int match_string(vector<char>& main, vector<char> substr){
vector<char>::iterator p,q, r;
int i = 0;
p = main.begin();
while (r <= substr.end() && p <= main.end()){
r = substr.begin();
while (*p != *r && p < main.end()){
p++;
i++;
}
q = p;
while (*p == *r && r <= substr.end() && p<=main.end()){
p++;
i++;
r++;
}
if (r >= substr.end()){
main.erase(main.begin(), q + 1);
return (i - substr.size() + 1);
}
if (p >= main.end())
return 0;
p = ++q;
}
}输出结果
Enter main String: C++ is programming language. It is object oriented language Enter substring to find: language Match found at Position = 20 Match found at Position = 52