题解 | #统计字符串中子串出现的次数#

#include <iostream>
#include <cstring>
using namespace std;

int main() {

    char str[100] = { 0 };
    char substr[100] = { 0 };

    cin.getline(str, sizeof(str));
    cin.getline(substr, sizeof(substr));

    int count = 0;

    // write your code here......
    // 双指针法
    char* p = str;
    char* q = substr;
    while(*p != '\0')
    {
        if(*p == *q)
        {
            //判断两个字符串同时结束的情况
            if(*(p + 1) == '\0' && *(q + 1) == '\0') count++;
            p++;
            q++;
        }
        if(*p != *q)
        { 
            if(*q == '\0')
            {
                count++;
                p -= q - substr - 1; //p也回溯，解决重叠问题
            }
            else p++;   
            q = substr;
        } 
    }


    cout << count << endl;

    return 0;
}