3Find duplicate codes |Huawei od computer test

Original question link: [Full marks][Huawei OD machine test real questions 2023 JAVA] Find duplicate codes_Ruo Bodou's blog-CSDN blog

c++ code:

#include <iostream>

using namespace std;

int main()
{
    string a;//0-200 暴力枚举
    string b;
    getline(cin, a);
    getline(cin, b);
    
    int len1 = a.length();
    int len2 = b.length();
    if(len1 > len2) swap(a, b);//让a是短的
    
    string res = "";
    
    //双指针 [i,j] 判断 n * n = 10000
    for(int j = len1 - 1; j > 0; j -- )
    {
        for(int i = 0; i < j; i ++ )
        {
            string temp = a.substr(i, j - i + 1);
            if(b.find(temp) != -1) {
                if(temp.length() > res.length())
                res = temp;
            }
        }
    }
    
    cout << res << endl;
    
    return 0;
}

Time complexity: O(n)

Because the string is 100 in length, it can be violently enumerated

Double pointers, time complexity n * n

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Origin blog.csdn.net/weixin_65293439/article/details/129835308