最长的公共子串

char base1[9]={'1','A','B','2','3','4','5','C','D'};
char base2[7]={'1','2','3','4','5','E','F'};
int longestCountinue()
{
    int dp[9][7]={{0},{0}};
    int i,j;
    for(i=0;i<9;i++)
    {
        if(base1[i] == base2[0])
        {
            dp[i][0] = 1;
        }
    }

    for(j=1;j<7;j++)
    {
        if(base2[j] == base1[0])
        {
            dp[0][j] = 1;
        }
    }

    /*dp[i][j]的含义是以base[i]和base2[j]为结尾的最大公共子串*/
    for(i=1;i<9;i++)
    {
        for(j=1;j<7;j++)
        {     
            if(base1[i] == base2[j])
            {
                dp[i][j] = dp[i-1][j-1] +1;
            }
        }
    }

    for(i=0;i<9;i++)
    {
        printf("\r\n");
        for(j=0;j<7;j++)
        {
            printf("%d ",dp[i][j]);
        }
    }
    /*找到最大值的下标*/
    int max = dp[0][0];
    int m = 0,n=0;
    printf("\r\n");
    for(i=0;i<9;i++)
    {
        for(j=0;j<7;j++)
        {
              if(dp[i][j] > max)
              {
                  max = dp[i][j];
                  m=i;
                  n=j;
                  printf("%d ",max);
              }
        }
    }
     printf("\r\nm:%d.n:%d ",m,n);
     i=m;
     int index = 4;
     char res[4] ={0};
     while(i>=0)
     {
         res[--index] = base1[i--];
     }
     char *p1;
     p1 = res;
     printf("%s",p1);
    return 0;
}
/*
 1 0 0 0 0 0 0 
 0 0 0 0 0 0 0 
 0 0 0 0 0 0 0 
 0 1 0 0 0 0 0 
 0 0 2 0 0 0 0 
 0 0 0 3 0 0 0 
 0 0 0 0 4 0 0 
 0 0 0 0 0 0 0 
 0 0 0 0 0 0 0 
 2 3 4 
 m:6.n:4 
 2345
*/
int longestCountinue2()
{
    int i,j,m=0,n=0;
    int row = 9;
    int len = 0;
    int max = 0;
    i=0;
    j=6;//定位到最右上角
    int f;
    while(i<row)
    {
        m = i;
        n=j;
        while(m<9 && n<7)
        {
            if(base1[m] == base2[n])
            {
                len ++;
                if(max<len)
                {
                    max =len;
                    f = m;
                    printf("max :%d ",max);
                }
            }
            else
            {
                len = 0;
            }
            m++;
            n++;

        }
        if(j>0)
        {
            j--;
        }
        if(j==0)
        {
            i++;
        }
    }
    int res = f-max + 1;
    printf("\r\n");
    for(i=0;i<max;i++)
    {
        printf("%c ",base1[res++]);
    }       
    return 0;
}