我有一个必须执行以下操作的方法:
for (int a01 = 1; a01 <= 25; a01++) {
for (int a02 = a01 + 1; a02 <= 25; a02++) {
for (int a03 = a02 + 1; a03 <= 25; a03++) {
...
System.out.println(a01 + "," + a02 + "," + ... + "," + a015);
}
}
}
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我想指定嵌套的数量(在上面的例子中,我想要15嵌套).有没有办法在这里使用递归编程?
Chr*_*ies 17
是.这可以通过递归编程来执行.
我假设你不喜欢在源代码中嵌套这些嵌套 - 就像在你的例子中一样,因为这是非常难看的编程 - 就像评论员解释的那样.
以下(伪Java类)代码说明了它.我假设嵌套的固定深度.那么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];
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必须将数组counters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.
我假设你喜欢在内循环中执行某个操作.我会称之为
performOperation(int[] counters);
- 这取决于多维计数器,即外部的计数器.
然后,您可以通过调用运行嵌套的for循环
nestedLoopOperation(counters, length, 0);
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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);
}
}
}
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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);
}
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