所以我尝试了这个程序，它应该找出这些门的位置。它们开始关闭，然后你切换每扇门。下次你每秒切换一次，然后每隔三，四，等等。我得到了正确的编码，但它似乎效率很低。如果我用更大的数字尝试这个，那么程序将花费太长时间。有没有办法更有效地做到这一点？这是我的编码：
public class doors
{
public static void main(String [] args)
{
int x, y = 1;
boolean [] doors = new boolean[100];
for (int a = 0; a < doors.length; a++)
doors[a] = false; //false means closed
do
{
for (x = 0; x < doors.length - 1; x++)
{
if ((x+1) % y == 0)
{
if (doors[x] == true)
doors[x] = false;
else if (doors[x] == false)
doors[x] = true;
}
}
y++;
}while (y <= doors.length);
for (boolean b : doors)
System.out.println(b);
}
}
// Thanks!!!
答案 0 :(得分：2)
让我们一步一步地简化代码。
1）分配数组时，所有元素都会初始化为其默认值，即0，false或null。在您的情况下，这意味着false，因此无需循环并执行此操作。
// before
boolean [] doors = new boolean[100];
for (int a = 0; a < doors.length; a++)
doors[a] = false; //false means closed
// after
boolean[] doors = new boolean[100];
2a）由于数组不是零长度数组，do-while循环也可能是正常的while循环（它至少会执行一次）。
2b）作为一个普通的while循环，在循环之前有一个初始化程序，在循环结束时有一个增量，它与for循环相同。
2c）由于循环后未使用y，因此可以在循环中声明它。
// before
int x, y = 1;
do
{
//code
y++;
}while (y <= doors.length);
// after
int x;
for (int y = 1; y <= doors.length; y++)
{
//code
}
3a）doors[x] == true与doors[x]相同。 doors[x] == false与! doors[x]相同。
3b）if (x) {} else if (! x) {}与if (x) {} else {}相同。 else之后的测试是多余的。
3c）if (x) x = false; else x = true;与x = ! x;相同。
3d）您可以通过^= true（XOR）跳过双重评估。
// before
if (doors[x] == true)
doors[x] = false;
else if (doors[x] == false)
doors[x] = true;
// after
doors[x] = ! doors[x];
// skip double evaluation of index lookup
doors[x] ^= true;
4a）(x+1) % y == 0仅适用于x，y-1，2*y-1的{{1}}值，依此类推，因此make loop start at { {1}}并按3*y-1递增，无需y-1语句。
4c）由于循环后未使用y，因此可以在循环中声明它。
if到目前为止的结果（在删除不必要的大括号后）
x 5a）将// before
int x;
for (x = 0; x < doors.length - 1; x++)
{
if ((x+1) % y == 0)
{
//code
}
}
// after
for (int x = y - 1; x < doors.length - 1; x += y)
{
//code
}
限制为boolean[] doors = new boolean[100];
for (int y = 1; y <= doors.length; y++)
for (int x = y - 1; x < doors.length - 1; x += y)
doors[x] ^= true;
for (boolean b : doors)
System.out.println(b);
意味着永远不会使用/更新数组的最后一个值。放下x。
< doors.length - 1 6a）每行打印一个布尔- 1 / // before
for (int x = y - 1; x < doors.length - 1; x += y)
// after
for (int x = y - 1; x < doors.length; x += y)
值的数组并不是人类可读的。打印在一行上作为二进制数。
true7a）在外环内移动打印，你会看到一个有趣的图案。
false// before
for (boolean b : doors)
System.out.println(b);
// after
for (boolean b : doors)
System.out.print(b ? '1' : '0');
System.out.println();
现在，当您查看结果模式（最后一行）时，您会看到您的代码可以缩减为（非常快）：
boolean[] doors = new boolean[100];
for (int y = 1; y <= doors.length; y++) {
for (int x = y - 1; x < doors.length; x += y)
doors[x] ^= true;
for (boolean b : doors)
System.out.print(b ? '1' : '0');
System.out.println();
}
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1001000010000001000000001000000000010000000000001000000000000001000000001111111101111111111111111110
1001000010000001000000001000000000010000000000001000000000000001000000000111111101111111111111111110
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1001000010000001000000001000000000010000000000001000000000000001000000000001111101111111111111111110
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1001000010000001000000001000000000010000000000001000000000000001000000000000011101111111111111111110
1001000010000001000000001000000000010000000000001000000000000001000000000000001101111111111111111110
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1001000010000001000000001000000000010000000000001000000000000001000000000000000001111111111111111110
1001000010000001000000001000000000010000000000001000000000000001000000000000000011111111111111111110
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答案 1 :(得分：1)
首先，doors数组中的默认值为false（但如果需要，可以使用Arrays.fill填充它）。接下来，您只需要将循环增加y。然后，您可以使用doors[x] = !doors[x];切换。像，
int y = 1;
boolean[] doors = new boolean[100];
// Arrays.fill(doors, false);
do {
for (int x = y - 1; x < doors.length; x += y) {
doors[x] = !doors[x];
}
y++;
} while (y <= doors.length);
System.out.println(Arrays.toString(doors));
或使用嵌套的for循环，例如
boolean[] doors = new boolean[100];
// Arrays.fill(doors, false);
for (int y = 1; y < doors.length; y++) {
for (int x = y - 1; x <= doors.length; x += y) {
doors[x] = !doors[x];
}
}
System.out.println(Arrays.toString(doors));
答案 2 :(得分：0)
正如上面提到的那样，你可以做很多事情来简化代码。通过这样做，您将开始更好地查看正在使用的算法，从而了解它们的改进位置。
使用这样的代码，简化像if语句这样的事情通常更多的是澄清代码而不是提高性能。我之所以这么说的原因是编译器这些天非常好，因为优化代码虽然可能会丢失一些指令，但性能差异可能很小。
如果有循环，你更有可能获得明显改善的地方。减少循环中的迭代次数通常会对总体时间产生显着影响。为此，我可以看到两个可以改进的地方。
首先，可能不需要初始设置循环。我没有检查过，但默认情况下门阵列可能是假的，因此可能会丢弃循环到初始化。
其次，内部循环递增1并测试该门是否与y匹配。你可以通过简单地用y增加内部循环来短路很多。这样你就会一直在想要翻转的门上。这意味着你可以放弃正确门的测试，然后翻转它。随着y的增加，x循环的迭代次数将减少。这意味着每次y都会加速。
最后，当我看到这样的代码时，我总是推荐一件事。始终在if语句中加上括号。当不使用括号时，很容易引入细微的错误。
答案 3 :(得分：-6)
而不是使用for语句，请使用foreach并切换
示例
for(boolean i:doors){
//do something
}