最长公共子串（LCS）

来源：互联网发布：linux autofs 编辑：程序博客网时间：2024/06/03 22:46

(1) 找出两个字符串的最长公共子串

题目：输入两个字符串，找出两个字符串中最长的公共子串。

找两个字符串的最长公共子串，这个子串要求在原字符串中是连续的。

#include<iostream>#include<string>using namespace std;int main(){    string s1 = "GCCCTAGCCAGDE";    string s2 = "GCGCCAGTGDE ";    int len1 = s1.size();    int len2 = s2.size();    int** c = new int* [len1];    for (int i = 0; i < len1; i++)    {        c[i] = new int[len2];        memset(c[i], 0, sizeof(int)*len2);    }    int start = -1;    int end = -1;    int maxLen = 0;    for (int i = 0; i < len1; i++)    {        for (int j = 0; j < len2; j++)        {            if (s1[i] == s2[j])            {                int indexI = i == 0 ? 0 : (i - 1);                int indexJ = j == 0 ? 0 : (j - 1);                c[i][j] = c[indexI][indexJ] + 1;                if (c[i][j] > maxLen)                {                    maxLen = c[i][j];                    end = i;                    start = i - c[i][j] + 1;                }            }            else            {                c[i][j] = 0;            }        }    }    if (maxLen != 0)    {        string res = s1.substr(start, maxLen);        cout << res << endl;    }    for (int i = 0; i < len1; i++)    {        delete[] c[i];    }    return 0;}

(2) 找出两个字符串的最长公共子序列

题目：输入两个字符串，求两个字符串的最长公共子序列。

首先，最长公共子序列与最长公共子串不同，子序列不要求其在原字符串是连续的。例如字符串X={A,B,C,B,D,A,B}，Y={B,D,C,A,B,A}，则X与Y的最长公共子序列为Z={B,C,B,A};

#include<iostream>#include<string>using namespace std;string getLongestSequence(string s1, string s2){    int len1 = s1.size();    int len2 = s2.size();    int** c = new int* [len1];    for (int i = 0; i < len1; i++)    {        c[i] = new int[len2];        memset(c[i], 0, sizeof(int)*len2);    }    int start = -1;    int end = -1;    int maxLen = 0;    for (int i = 0; i < len1; i++)    {        for (int j = 0; j < len2; j++)        {            if (s1[i] == s2[j])            {                int preMax = (i == 0 || j == 0) ? 0 : c[i - 1][j - 1];                c[i][j] = preMax + 1;            }            else            {                int preMax1 = (i == 0) ? 0 : c[i - 1][j];                int preMax2 = (j == 0) ? 0 : c[i][j - 1];                c[i][j] = max(preMax1, preMax2);            }            if (c[i][j] > maxLen)            {                maxLen = c[i][j];            }        }    }    string res(maxLen, '-');    for (int i = len1 - 1, j = len2 - 1, index = maxLen; i >= 0 && j >= 0;)    {        int preC1 = i == 0 ? -1 : c[i - 1][j];        //说明s1[i]字符不在最大子序列里        if (c[i][j] == preC1)        {            i--;            continue;        }        //说明s1[i]字符在最大子序列里        else        {            //更新结果            res[--index] = s1[i];            //说明s2[j]字符不在最大子序列里            int preC2 = j == 0 ? -1 : c[i][j - 1];            while (c[i][j] == preC2)            {                j--;            }            i--;            j--;        }    }    return res;    for (int i = 0; i < len1; i++)    {        delete[] c[i];    }}int main(){    string s1 = "ABCBDAB";    string s2 = "BDCABA";    string res1 = getLongestSequence(s1, s2);    string res2 = getLongestSequence(s2, s1);    if (strcmp(res1.c_str(), res2.c_str()))    {        cout << res1 << endl;        cout << res2 << endl;    }    else    {        cout << res1 << endl;    }    return 0;}

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