java 数组中某个数出现的概率_java – 如何为数组中的某个重复数字获取一组频率？...

以下程序使用具有开始和结束的规则.允许规则重叠,以便可以在2个或更多规则中使用1.如果这不是您想要的,那么修改代码将非常容易.

如果1的数量很小,它应该非常快,但它不能很好地扩展.

它不是很“聪明”.它所做的只是尽可能在每个阶段淘汰尽可能多的人.这种方法不是最优的,但在大多数情况下应该非常好.例如,如果你开始

1 1 1 1 1 0 1 1 1 1 1

找到的第一条规则是

Every 2th, starting at 1, ending at 11.

因为它使用了六个.但是,最好的解决方案显然只需要两个涉及五个1的规则.我认为找到一个始终能找到最佳答案的有效算法是非常困难的.

public static void main(String[] args) {

rulesFor(0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0);

}

public static void rulesFor(int... arr) {

Set allOnes = new HashSet<>();

for (int i = 0; i < arr.length; i++)

if (arr[i] == 1)

allOnes.add(i);

// Size 1 has to be done separately as the below code wouldn't work.

if (allOnes.size() == 1) {

int a = allOnes.iterator().next();

System.out.println("Every 1st, starting at " + (a + 1) + ", ending at " + (a + 1) + ".");

return;

}

Set leftToRemove = new HashSet<>(allOnes);

while (!leftToRemove.isEmpty()) {

int low = -1;

int high = -1;

int d = -1;

int removeTotal = -1;

for (int a : leftToRemove) {

for (int b : allOnes) {

if (b == a)

continue;

int d2 = Math.abs(b - a);

int low2 = a;

int high2 = a;

int removeTotal2 = 1;

while (true) {

if (!allOnes.contains(low2 - d2))

break;

low2 -= d2;

if (leftToRemove.contains(low2))

removeTotal2++;

}

while (true) {

if (!allOnes.contains(high2 + d2))

break;

high2 += d2;

if (leftToRemove.contains(high2))

removeTotal2++;

}

if (removeTotal2 > removeTotal) {

low = low2;

high = high2;

removeTotal = removeTotal2;

d = d2;

}

}

}

System.out.println("Every " + d + "th, starting at " + (low + 1) + ", ending at " + (high + 1) + ".");

for (int i = low; i <= high; i += d)

leftToRemove.remove(i);

}

}