#include<iostream>
#include<string>
#include<vector>
#include<math.h>
using namespace std;
//从按行的角度出发,计算逐个需要存的字符的对应原串的下标。
class Solution {
public:
int strStr(string haystack,string needle) {
int i,j;
// cout << needle.size();
if (needle.size() > haystack.size())
return -1;
for (i = 0; i < haystack.size() - needle.size()+1; i++){
for (j = 0; j < needle.size(); j++)
if (haystack[i + j] != needle[j]) {
break;
}
if (j == needle.size()){
return i;
}
}
return -1;
}
};
int main(){
Solution exp;
string str ="abb";
string str2 = "abaaa";
int re =exp.strStr(str,str2);
cout << re;
system("pause");
return 1;
}
strStr()
最新推荐文章于 2025-03-22 08:57:15 发布