Question

我正在处理一个名为Types and Tags的两个对象的应用程序。

Type.java

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <ctype.h>

#define MAX_LEN 10
#define SPECIAL '#'

bool f[MAX_LEN+1][MAX_LEN+1][MAX_LEN+1];

bool isSpecialName(char *pa, char *pb, char *pc) {
    int la = strlen(pa);
    int lb = strlen(pb);
    int lc = strlen(pc);

    if (la > lb + lc) return false;

    memset(f, false, sizeof(f));
    memset(f[0], true, sizeof(f[0]));

    for (int i=1; i<=la; ++i) for (int j=0; j<=lb; ++j) for (int k=0; k<=lc; ++k) {
        char a = tolower(pa[i-1]);
        char b = j > 0 ? tolower(pb[j-1]) : SPECIAL;
        char c = k > 0 ? tolower(pc[k-1]) : SPECIAL;

        if (j > 0)  f[i][j][k] = f[i][j-1][k]   || f[i][j][k];
        if (k > 0)  f[i][j][k] = f[i][j][k-1]   || f[i][j][k];
        if (a == b) f[i][j][k] = f[i-1][j-1][k] || f[i][j][k];
        if (a == c) f[i][j][k] = f[i-1][j][k-1] || f[i][j][k];
    }

    return f[la][lb][lc];
}

void check(char *a, char *b, char *c) {
    if (isSpecialName(a, b, c)) fprintf(stdout, "'%s' *IS* special name of '%s' and '%s'\n", a, b, c);
    else fprintf(stderr, "'%s' is *NOT* special of '%s' and '%s'\n", a, b, c);
}

int main() {
    check("ab", "a", "b");
    check("Dan", "Jane", "Adam");
    check("Henry", "Hunter", "Wendy");
    check("abcd", "ac", "bd");
    check("abcd", "ac", "bb");
    return 0;
}

Tag.java

public class Type {
    private String name;
    private final UUID uuid;
    private final List<Tag> tags;
}

Tables.sql

public class Tag {
    private String name;
    private final UUID uuid;
}

我遵循DAO设计模式，但我在这里应用它时遇到了麻烦。每个Type都有一个Tags列表，那么加载Type的最佳方法是什么？

DAO模式与列表

0 个答案: