Question

我创建了一个脚本，根据所选的选项显示结果。

我使用：A，B，C，D，E，F，G，H（将来会更多）

int A = 0;
int B = 0;
int C = 0;
int D = 0;
int E = 0;
int F = 0;
int G = 0;
int H = 0;

最终的ansver将是16个组合中的一个。

A;C;E;G = Nr1
A;C;E;H = Nr2
A;C;F;G = Nr3
A;C;F;H = Nr4
A;D;E;G = Nr5
A;D;E;H = Nr6
A;D;F;G = Nr7
A;D;F;H = Nr8
B;C;E;G = Nr9
B;C;E;H = Nr10
B;C;F;G = Nr11
B;C;F;H = Nr12
B;D;E;G = Nr13
B;D;E;H = Nr14
B;D;F;G = Nr15
B;D;F;H = Nr16

我想展示潜在的变种按下组合时：A, C' then answer is '"Nr1,Nr2,Nr3,Nr4"
按下组合时：A, G' then answer is '"Nr1,Nr3,Nr5,Nr7"
稍后会有更多变量I，J，K，L ....等等。但答案将只有第16个
A数据结构的逻辑可能是什么，例如Map，我有点卡住了？

重要 - 可以创建组合混合案例例如：H;C;E;A或E;C;A;H ....等答案为Nr1

If / Else似乎太长了。目前代码：

    String scoreTeamA = "waiting for the results";


    if (A == 1) {
        if (C == 1) {
            if (E == 1) {
                if (G == 1) {
                    scoreTeamA = "The answer is: Nr1"; //combination: A;C;E;G
                } else if (H == 1) {
                    scoreTeamA = "The answer is: Nr2"; //combination: A;C;E;H
                } else scoreTeamA = "Possible variants, one of: Nr1, Nr2"; //combination: A;C;E

            } else if (F == 1) {
                if (G == 1) {
                    scoreTeamA = "The answer is: Nr3"; //combination: A;C;F;G
                } else if (H == 1) {
                    scoreTeamA = "The answer is: Nr4"; //combination: A;C;F;H
                } else scoreTeamA = "Possible variants, one of: Nr3,Nr4"; //combination: A;C;F

            } else scoreTeamA = "Possible variants, one of: Nr1,Nr2,Nr3,Nr4"; //combination: A;C;

        } else if (D == 1) {
            scoreTeamA = "Possible variants, one of: Nr5,Nr6,Nr7,Nr8";//combination: A;D;
        } else
            scoreTeamA = "Possible variants, one of: Nr1,Nr2,Nr3,Nr4,Nr5,Nr6,Nr7,Nr8"; //combination: A
    } else if (B == 1) { 
        scoreTeamA = "Possible variants, one of: Nr9,Nr10,Nr11,Nr12,Nr13,Nr14,Nr15,Nr16"; //combination: B
    }

Answer 1

创建POJO。布尔值仍然有效，因为您可以选择null。选择a时，a为真。选择b时，b为假。没有选择为空。

class Option {
    Boolean a;
    Boolean c;
    Boolean e;
    Boolean g;
    String value;

    //do some getters and setters

    boolean isValid(Boolean a, Boolean c, Boolean e, Boolean g) {
        if(this.a != a && a != null) {
            return false;
        }
        if(this.c != c && c != null) {
            return false;
        }
        if(this.e != e && e != null) {
            return false;
        }
        if(this.g != g && g != null) {
            return false;
        }

        return true;
    }
}

我们可以使用此填充我们的选项 https://stackoverflow.com/a/27008250/2958086

List<Option> option = new ArrayList<>();
final int n = 4;
for (int i = 0; i < Math.pow(2, n); i++) {
    String bin = Integer.toBinaryString(i);
    while (bin.length() < n)
        bin = "0" + bin;
    char[] chars = bin.toCharArray();
    boolean[] boolArray = new boolean[n];
    for (int j = 0; j < chars.length; j++) {
        boolArray[j] = chars[j] == '0' ? true : false;
    }
    list.add(new Option(boolArray, i+1)); //create this constructor
}

然后，搜索：

List<String> findValues(Boolean a, Boolean c, Boolean e, Boolean g) {
    List<String> values = new ArrayList<>();

    for(Option option : options) {
        if(option.isValid(a, c, e, g))
            values.add(option.getValue());
    }
    return values;      
}

使用：

List<String> myValues = findValues(true, false, null, null); // selects all filings that are AD**

任何填充为null的值都将通过，任何不会被计算的值都会与它们各自的值进行比较。

将if / else更改为短地图逻辑

1 个答案: