假设我有一个列表,每个列表都包含一个'would be‘for循环,我无法理解下面这一点。每一次继承都应该在彼此之间。
因此,如果我有一个包含3个对象的列表,我想
class Focus {
String focus;
List<String> values;
public Focus(String focus, String... values) {
this.focus = focus;
this.values = Lists.newArrayList(values);
}
}
List<Focus> focuses = new ArrayList<Focus>();
focuses.add(new Focus("Focus 1", "09", "14", "13", "12"));
focuses.add(new Focus("Focus 2", "94", "92"));
focuses.add(new Focus("Focus 3", "A", "B"));
String my_string = "";
for (Focus obj1 : list_obj_x) {
for (Focus obj2 : list_obj_xx) {
for (Focus obj3 : list_obj_xxx) {
my_string += obj1 + " " + obj2 + " " + obj3;
}
}
}显然,使用list时,for循环结构可能会增长,而上述情况是不可能的。我需要一个动态的结构来满足my_string的需要。即:
94 09 A
94 14 A
94 13 A
94 12 A
94 09 B
94 14 B
94 13 B
94 12 B
92 09 A
92 14 A
92 13 A
92 12 A
92 09 B
92 14 B
92 13 B
92 12 B 输出应该类似于上面的内容。这就是我到目前为止所知道的:
int focusCount = focuses.size();
for (int i = (focusCount - 1); i >= 0; i--) {
Focus currentFocus = focuses.get(i);
List<String> currentFocusValues = currentFocus.values;
for (int cfv = 0; cfv < currentFocusValues.size(); cfv++) {
String currentFocusValue = currentFocusValues.get(cfv);
for (int j = (i - 1); j >= 0; j--) {
Focus previousFocus = focuses.get(j);
List<String> previousFocusValues = previousFocus.values;
for (int pfv = 0; pfv < previousFocusValues.size(); pfv++) {
String previousFocusValue = previousFocusValues.get(pfv);
System.out.println(currentFocusValue + " " + previousFocusValue);
}
}
}
}它迎合了列表值的所有组合,但不是我想要的结构。
有人能帮帮我吗?
发布于 2012-02-03 21:23:17
最直接的方法可能是递归。在递归的每个步骤中,您逐个“锁定”第n个列表的值,然后向下递归“列表列表”,直到您到达末尾。
String[] values = new String[focuses.size()];
CreateCombinations(focuses, 0, values);使用递归方法
private void CreateCombinations(List<Focus> focuses, int index, string[] values) {
Focus focus = focuses.get(index);
for (string v : focus.values) {
values[index] = v;
if (index < focuses.size() - 1) {
// there is at least one other focus
CreateCombinations(focuses, index+1, values);
} else {
// all values pinned down
StringBuilder sb = new StringBuilder(values[0]);
for (int i = 1; i < values.length; ++i) {
sb.append(" ").append(values[i]);
}
// now do whatever you like to do with sb.toString()...
}
}
}当然,这可以进一步改进,但对于您来说,这可能足以作为一个起点。
发布于 2012-02-03 22:16:12
这里有一个迭代的方法(仍然需要清理):
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Focus {
String focus;
List<String> values;
public Focus(String focus, String... values) {
this.focus = focus;
this.values = Arrays.asList(values);
}
public static String printAllCombinations(Focus... focuses) {
String myString = "";
List<String> allCombinations = new ArrayList<String>();
int length = focuses.length;
if (length == 0) {
return "";
} else if (length == 1) {
allCombinations = focuses[0].values;
} else if (length > 1) {
for (Focus f : focuses) {
allCombinations = getCombinations(allCombinations, f.values);
}
}
for (String s : allCombinations) {
myString += s+"\n";
}
return myString;
}
private static List<String> getCombinations(List<String> l1, List<String> l2) {
if (l1.size() == 0) {return l2;}
else if (l2.size() == 0) {return l1;}
List<String> combinations = new ArrayList<String>();
for (String outerValue : l1) {
for (String innerValue : l2) {
combinations.add(outerValue + " " + innerValue);
}
}
return combinations;
}
}发布于 2012-02-03 21:09:01
您只有两个循环:一个用于焦点对象,一个用于它们的值。
https://stackoverflow.com/questions/9129116
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