【编程题目】最长公共字串

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void printLCS(char * X, int m, int n, int i, int j, char * B) 
{
    if (i < 0 || j < 0)
    {
        return;
    }
    char b = *(B + i * n + j);
    if (b == ‘y‘)
    {
        printLCS(X, m , n, i - 1, j - 1, B);
        printf("%c ", X[i]); //先得到的是后面的共同字母，应该在后面打印出来
    }
    else if (b == ‘u‘)
    {
        printLCS(X, m , n, i - 1, j, B);
    }
    else
    {
        printLCS(X, m , n, i, j - 1, B);
    }
}

int LCS(char * X, char * Y, int m, int n, char * & B) //返回最大公共子串的长度
{
    int * C = (int *)malloc((m + 1) * (n + 1) * sizeof(int)); //记录每个子问题的最长公共子序列 0-m 0-n
    B = (char *)malloc(m * n * sizeof(char)); //标记 1-m 1-n
    memset(C, 0, (m + 1) * (n + 1) * sizeof(int)); //初始化为全0

    for (int i = 0; i < m; i++)
    {
        int * crow_1 = C + i * (n + 1); //注意 C 中 i = 1 j = 1 对应 X的X[0] 与 Y的Y[0]
        int * crow = C + (i + 1) * (n + 1);
        char * brow = B + i * n;  //注意 B 中 i = 0 j = 0 对应 X的X[0] 与 Y的Y[0]
        for(int j = 0; j < n; j++)
        {
            if (X[i] == Y[j])
            {
                crow[j + 1] = crow_1[j] + 1;
                brow[j] = ‘y‘;
            }
            else
            {
                crow[j + 1] = (crow_1[j + 1] > crow[j]) ? crow_1[j + 1] : crow[j];
                brow[j] = (crow_1[j + 1] > crow[j]) ? ‘u‘ : ‘l‘;
            }
        }
    }

    printLCS(X, m, n, m - 1, n - 1, B);
    int maxlen = *(C + m * (n + 1) + n);
    free(C);
    free(B);
    return maxlen;
}

int main()
{
    char X[7] = {‘A‘,‘B‘,‘C‘,‘B‘,‘D‘,‘A‘,‘B‘};
    char Y[6] = {‘B‘,‘D‘,‘C‘,‘A‘,‘B‘,‘A‘};
    char * B = NULL;
    int l = LCS(X, Y, 7, 6, B);

    return 0;
}