DS--最长重复子串

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题目描述

求串的最长重复子串长度（子串不重叠）。例如：abcaefabcabc的最长重复子串是串abca，长度为4。

输入

测试次数t

t个测试串

输出

对每个测试串，输出最长重复子串长度，若没有重复子串，输出-1.

样例输入

复制代码

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3
abcaefabcabc
szu0123szu
szuabcefg

样例输出

复制代码

思路：

实现代码：

复制代码

#include<iostream>
#include<string>
 
using namespace std;
 
void GetNext(string p, int next[])
{
    next[0] = -1;
    int i, j;
    i = 0;
    j = -1;
    while (i<p.length())
    {
        if (j == -1 || p[i] == p[j])
        {
            ++i;
            ++j;
            next[i] = j;
        }
        else
            j = next[j];
    }
}
 
int main()
{
    int n;
    cin >> n;
    string s;
    string re;
    int i, j = -1,k;
    int max = 0;
    while (n--)
    {
        int next[100] = { 0 };
        max = -1;
        cin >> s;
        re=s;
        for (i = 0, k = s.length() - 1; k >= 0; i++, k--)
        {
            re[i] = s[k];
        }
        GetNext(s, next);         
        for (i = 0; i <= s.length(); ++i)
        {
            if (next[i]>max && next[i]<=s.length()/2)
                max = next[i];
        }   
        GetNext(re, next);
        for (i = 0; i <= re.length(); ++i)
        {
            if (next[i]>max && next[i]<=re.length()/2)
                max = next[i];      
        }         
        if (max == 0)
            cout << j << endl;
        else
        {
            cout << max << endl;
        } 
    }
    return 0;
}

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#include<iostream>
#include<string>
 
using namespace std;
 
void GetNext(string p, int next[])
{
    next[0] = -1;
    int i, j;
    i = 0;
    j = -1;
    while (i<p.length())
    {
        if (j == -1 || p[i] == p[j])
        {
            ++i;
            ++j;
            next[i] = j;
        }
        else
            j = next[j];
    }
}
 
int main()
{
    int n;
    cin >> n;
    string s;
    string re;
    int i, j = -1,k;
    int max = 0;
    while (n--)
    {
        int next[100] = { 0 };
        max = -1;
        cin >> s;
        re=s;
        for (i = 0, k = s.length() - 1; k >= 0; i++, k--)
        {
            re[i] = s[k];
        }
        GetNext(s, next);         
        for (i = 0; i <= s.length(); ++i)
        {
            if (next[i]>max && next[i]<=s.length()/2)
                max = next[i];
        }   
        GetNext(re, next);
        for (i = 0; i <= re.length(); ++i)
        {
            if (next[i]>max && next[i]<=re.length()/2)
                max = next[i];      
        }         
        if (max == 0)
            cout << j << endl;
        else
        {
            cout << max << endl;
        } 
    }
    return 0;
}