是的。这可以通过递归编程来执行。
我假设您不喜欢在源代码中写下这些嵌套的 for - 就像在您的示例中一样，因为这确实是丑陋的编程 - 就像评论者解释的那样。
以下（伪 Java 类）代码说明了这一点。我假设嵌套的深度是固定的。然后你实际上喜欢循环一个维度深度的整数向量。
int[] length = new int[depth];
int[] counters = new int[depth];
数组counters必须初始化为 0 ( Arrays.fill(counters,0))。必须将数组length初始化为相应 for 循环的迭代次数。
我假设您喜欢在内部循环中执行某个操作。我会称之为
performOperation(int[] counters);
- 它取决于多维计数器，即外部 for 的计数器。
然后您可以通过调用运行嵌套的 for 循环
nestedLoopOperation(counters, length, 0);
在哪里
void nestedLoopOperation(int[] counters, int[] length, int level) {
if(level == counters.length) performOperation(counters);
else {
for (counters[level] = 0; counters[level] < length[level]; counters[level]++) {
nestedLoopOperation(counters, length, level + 1);
}
}
}
在您的情况下，您的 System.out.println() 将是
performOperation(int[] counters) {
String counterAsString = "";
for (int level = 0; level < counters.length; level++) {
counterAsString = counterAsString + counters[level];
if (level < counters.length - 1) counterAsString = counterAsString + ",";
}
System.out.println(counterAsString);
}
我创建了这个程序来显示所有不同可能的卡片组合（不重复）。它使用递归 for 循环。也许它可以帮助你。
//I'm lazy, so yeah, I made this import...
import static java.lang.System.out;
class ListCombinations {
//Array containing the values of the cards
static Symbol[] cardValues = Symbol.values();
//Array to represent the positions of the cards,
//they will hold different card values as the program executes
static Symbol[] positions = new Symbol[cardValues.length];
//A simple counter to show the number of combinations
static int counter = 1;
/*Names of cards to combine, add as many as you want, but be careful, we're
talking about factorials here, so 4 cards = 24 different combinations (4! = 24),
but 8 cards = 40320 combinations and 13 cards = 6.23 billion combinations!*/
enum Symbol {
AofSpades, TwoofSpades, ThreeofSpades, FourofSpades
}
public static void main(String args[]) {
//I send an argument of 0 because that is the first location which
//we want to add value to. Every recursive call will then add +1 to the argument.
combinations(0);
}
static void combinations(int locNumber) {
/* I use a recursive (repeating itself) method, since nesting for loops inside
* each other looks nasty and also requires one to know how many cards we will
* combine. I used 4 cards so we could nest 4 for loops one after another, but
* as I said, that's nasty programming. And if you add more cards, you would
* need to nest more for loops. Using a recursive method looks better, and gives
* you the freedom to combine as many cards as you want without changing code. */
//Recursive for loop nesting to iterate through all possible card combinations
for(int valueNumber = 0; valueNumber < cardValues.length; valueNumber++) {
positions[locNumber] = cardValues[valueNumber];
if (locNumber < (cardValues.length-1)) {
combinations(locNumber + 1);
}
//This if statement grabs and displays card combinations in which no card value
// is repeated in the current "positions" array. Since in a single deck,
// there are no repeated cards. It also appends the combination count at the end.
if (locNumber == (cardValues.length-1) && repeatedCards(positions)) {
for (int i = 0; i < cardValues.length; i++) {
out.print(positions[i]);
out.print(" ");
}
out.printf("%s", counter);
counter++;
out.println();
}
}
}
static boolean repeatedCards(Symbol[] cards) {
/*Method used to check if any cards are repeated in the current "positions" array*/
boolean booleanValue = true;
for(int i = 0; i < cardValues.length; i++) {
for(int j = 0; j < cardValues.length; j++) {
if(i != j && cards[i] == cards[j]) {
booleanValue = false;
}
}
}
return booleanValue;
}
}