Question

我试图解决最常见的子序列问题。我面临整数数组到指针转换的问题。我的2D数组的最后一个元素将是结果。这是我的代码：

#include<iostream>
#include<cstring>
using namespace std;

char firstSequence_X[999999],secondSequence_Y[999999];
int length_M,length_N;


void LCS_LENGTH(char*,char*);

int main(void)
{
    LCS_LENGTH(firstSequence_X,secondSequence_Y);
    return 0;

}

void LCS_LENGTH(char X[],char Y[])
{
    //scan
    cout<<"Enter first sequence : ";
    cin>>X;
    length_M=strlen(X);
    cout<<"Enter second sequence : ";
    cin>>Y;
    length_N=strlen(Y);
    cout<<endl;
    cout<<"First sequence's length is : "<<length_M<<endl;
    cout<<"Second sequence's length is : "<<length_N<<endl;
    //declare 2 matrix : C for length , B for sign
    int C[length_M+1][length_N+1],B[length_M+1][length_N+1];
    //loop variable
    int i,j;
    //clear all
    for(i=0;i<=length_M;i++)
    {
        C[i][0]=0;
        B[i][0]='n';
    }
    for(j=0;j<=length_N;j++)
    {
        C[0][j]=0;
        B[0][j]='n';
    }
    //dp
    for(i=1;i<=length_M;i++)
    {
    for(j=1;j<=length_N;j++)
    {
        if(X[i]==Y[j])
        {
            C[i][j]=C[i-1,j-1]+1;//diagonal+1
            B[i][j]='d';
        }
        else if(C[i-1][j]>=C[i][j-1])
        {
            C[i][j]=C[i-1,j];//up row copy
            B[i][j]='u';
        }
        else
        {
            C[i][j]=C[i-1,j];//left column copy
            B[i][j]='l';
        }
    }
}
cout<<endl<<"C[i,j] :"<<endl;
for(i=0;i<=length_M;i++)
{
    for(j=0;j<=length_N;j++)
    {
        cout<<C[i][j]<<" ";
    }
    cout<<endl;
}
cout<<endl<<"B[i,j] :"<<endl;
for(i=0;i<=length_M;i++)
{
    for(j=0;j<=length_N;j++)
    {
        cout<<B[i][j]<<" ";
    }
    cout<<endl;
}
}

我必须打印B和C数组。但是显示了以下错误消息：从int *到int的无效转换。请纠正我的错误。 TIA

Answer 1

看起来你做错了以下事情。

WebUtilLogging=on
WebUtilLoggingDetail=normal

我认为这应该是

C[i][j]=C[i-1,j-1]+1;//diagonal+1

和

C[i][j]=C[i-1][j-1]+1;//diagonal+1

应该是

 C[i][j]=C[i-1,j];//up row copy

和

 C[i][j]=C[i-1][j];//up row copy

应该是

 C[i][j]=C[i-1,j];//left column copy

键入从指针到数组的转换

1 个答案: