我創建了一個腳本,根據所選選項顯示結果。更改long if/else以縮短地圖邏輯
我使用:A,B,C,d,E,F,G,H(在未來將會更)
int A = 0;
int B = 0;
int C = 0;
int D = 0;
int E = 0;
int F = 0;
int G = 0;
int H = 0;
最終ansver將16組合中的一個。
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
我想顯示潛在的變體。
當你按下的組合:A, C' then answer is '"Nr1,Nr2,Nr3,Nr4"
當你按下的組合:A, G' then answer is '"Nr1,Nr3,Nr5,Nr7"
以後會更多的變數I
,J
,K
,L
....等,但答案將只有16
什麼可能是一個數據結構的邏輯,如Map
,我有點卡住了?
重要 - 組合可以產生也和案件 例如:H;C;E;A
或E;C;A;H
....等答案是Nr1
If/Else
似乎太長。 當前的代碼:
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
}
我看到四個二進制變量。使用樹。 – Compass
@HypnicJerk和@Compass我可以縮短變量的數量;到4('A,B,C,D'),那麼我得到這些元素的每一個是3值'-1; 0; 1' –
也許一個['BitSet'](http://docs.oracle.com/javase/8/docs/api/java/util/BitSet.html)作爲這些字母的任意組合的表示將有所幫助。然後你可以使用一個Map來獲得答案。 –
Seelenvirtuose