對於尋找兩個字串的最長公共子字串的問題，暴力搜尋的方式的時間複雜度將高達O(n^3), 而通過字尾樹的方式可將時間複雜度降低到O(n^2)。

以下是我實現的C++原始碼：

#include <iostream>
#include <string>
#include <vector>

using namespace std;

class Solution
{
public:
    int FindLongestCommonSubstring(string str1, string str2, vector<string>& MaxSubstr)
    {
        int
 
 m = str1.length(), n = str2.length(), i, j, k = 0, MaxSubstrLen = 0;
        vector<vector<int>> L(m, vector<int>(n, 0));  // L記錄相同子串的長度

        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                if (str1[i] == str2[j])
                {
                    if
 
 (i == 0 || j == 0)
                        L[i][j] = 1;
                    else
                        L[i][j] = L[i - 1][j - 1] + 1;

                    if (MaxSubstrLen < L[i][j])
                    {
                        MaxSubstrLen = L[i][j];
                        k = 0;
                        MaxSubstr[k] = str1.substr(i + 1
 
 - MaxSubstrLen, MaxSubstrLen);
                    }
                    else if (MaxSubstrLen == L[i][j] && MaxSubstr[k] != str1.substr(i + 1 - MaxSubstrLen, MaxSubstrLen))  
                        //有多個長度相同的MaxSubstr, 且要排除多個MaxSubstr內容相同的情況，如當str1="a", str2="aaaa"
                        MaxSubstr[++k] = str1.substr(i + 1 - MaxSubstrLen, MaxSubstrLen);
                }
                else
                    L[i][j] = 0;
            }
        }

        return k;
    }
};

int main()
{
    string str1 = "aaabaaaeabcd";
    string str2 = "bbbababcde";
    vector<string> MaxSubstr(str1.length());  //MaxSubstr個數不會超過str1和str2的長度，這裡取str1.length
    Solution a;
    int MaxSubstrNum = a.FindLongestCommonSubstring(str1, str2, MaxSubstr);

    for (int i = 0; i <= MaxSubstrNum; i++)
        cout << MaxSubstr[i] << ' ';
    cout << endl;

    return 0;
}