Java 数组分区最小差
我需要根据C#/Java中的最小差异对数组数据进行分区 e、 gJava 数组分区最小差,java,c#,algorithm,Java,C#,Algorithm,我需要根据C#/Java中的最小差异对数组数据进行分区 e、 g B3和D4之间的差值=| 3-4 |=1 A23和B25之间的差值=| 23-25 |=2 D4和C7之间的差值=| 4-7 |=3 B12和A12之间的差值=| 12-12 |=0 这些规则是： 对于每组，字母不能重复 对于每个组，它可以包含1-4个元素 元素之间的差异必须您提供的输入包含多个解决方案。大概15岁左右（A1-A2、A35-A36、D4-C7将改变解决方案）。因为我问你想要哪种解决方案时你没有回答，所以我写了
- B3和D4之间的差值=| 3-4 |=1
- A23和B25之间的差值=| 23-25 |=2
- D4和C7之间的差值=| 4-7 |=3
- B12和A12之间的差值=| 12-12 |=0
- 对于每组，字母不能重复
- 对于每个组，它可以包含1-4个元素
- 元素之间的差异必须您提供的输入包含多个解决方案。大概15岁左右（A1-A2、A35-A36、D4-C7将改变解决方案）。因为我问你想要哪种解决方案时你没有回答，所以我写了这段代码，它将为这个问题提供一个（最简单的）“解决方案=）
在这种情况下，Solve（）的输出将是：
{ A2, B3, D4 },{ C7 },{ B12, A12, C14, D15 },{ C22, A23, B25 },{ A35, D37 },{ A36 }
重要提示：此代码假定输入中的字符串是唯一的。用并行数组处理语言编写代码只花了10分钟，但我花了2小时写下答案。为了不打破问题中的语言限制，我将不提交任何代码，但我将尝试使用数据和一些伪代码来澄清以下原则 如果参数具有固定顺序，则有512种可能的解决方案，如下所示：
具有最小分区（6）的解决方案是： 下一个（有7个分区）是： 最后一个（分成15个分区）自然是：┌──────────┬─┬─┬──┬──┬───┬───┬──┬──┬──┬──┬─────┐ │Partitions│6│7│8 │9 │10 │11 │12│13│14│15│Total│ ├──────────┼─┼─┼──┼──┼───┼───┼──┼──┼──┼──┼─────┤ │Solutions │1│9│36│84│126│126│84│36│9 │1 │512 │ └──────────┴─┴─┴──┴──┴───┴───┴──┴──┴──┴──┴─────┘
解决这个问题的最好、最安全的方法是使用蛮力并遍历所有可能性。首先，您需要调整使用0和1来指示分区位置的原则。因为参数中有15个元素，所以我们使用15个长度的二进制向量来完成这项工作。例如：┌──┬──┬──┬──┬──┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐ │A1│A2│B3│D4│C7│B12│A12│C14│D15│C22│A23│B25│A35│A36│D37│ └──┴──┴──┴──┴──┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┘
暗示/应返回4个分区：(1 0 0 1 0 0 0 0 1 0 0 1 0 0 0) partition 'stackoverflowex'
您还可以分割另一个15长度的布尔向量：┌───┬─────┬───┬────┐ │sta│ckove│rfl│owex│ └───┴─────┴───┴────┘
应返回：(1 0 0 1 0 0 0 0 1 0 0 1 0 0 0) partition (1 1 0 0 1 1 0 1 0 0 0 1 1 0 0)
您可以计算每个分区中的和。上面的一个将返回：┌─────┬─────────┬─────┬───────┐ │1 1 0│0 1 1 0 1│0 0 0│1 1 0 0│ └─────┴─────────┴─────┴───────┘
要解决问题，必须生成所有可能的分区向量。做起来比说起来简单。您只需要这两者之间的所有分区向量：┌─┬─┬─┬─┐ │2│3│0│2│ └─┴─┴─┴─┘
这些是什么？非常简单，它们是16384和32767的2基（二进制）表示。您必须简单地循环遍历16384和32767之间的所有数字（包括这两个数字），将每个数字转换为2进制，按此对数据进行分区，然后查看当前分区是否可接受（=符合您的标准，如“对于每个组，字母不能重复”）。转换间隔中的所有数字将覆盖所有可能的分区-任何可能的零和一的组合都在其中。计算只需要一秒钟的时间 伪：1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 // This would create 1 single, big partition 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 // This would create 15 small partitions// The character part of the argument: 15-length vector of characters: Chars = "A","A","B","D","C","B","A","C","D","C","A","B","A","A","D" // From that, extract the locations of the unique characters: CharsA = 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0 // Where Chars == A CharsB = 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0 // Where Chars == B CharsC = 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0 // Where Chars == C CharsD = 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1 // Where Chars == D // The numeric part of the argument: 15-length vector of numbers: // Btw, is this about lottery... hmmm Nums = 1, 2, 3, 4, 7, 12, 12, 14, 15, 22, 23, 25, 35, 36, 37 :For Number :In [all numbers between 16384 and 32767] Base2 = [2-base of Number] // 20123 would give: 1 0 0 1 1 1 0 1 0 0 1 1 0 1 1 // Test 1: "For each group, it can contains 1 - 4 elements" [using Base2, partition the partition vector Base2 itself; bail out if any partition length exceeds 4] // Test 2: "Difference between element must be <= 3" [using Base2, partition Nums; check differences inside each partition; bail out if bigger than 3 anywhere] // Test 3: "For each group, letter cannot be repeated" [using Base2, partition CharsA, CharsB, CharsC, CharsD (each in turn); count number of ones in each partition; bail out if exceeds 1 anywhere (meaning a character occurs more than once)] // If we still are here, this partition Number is ACCEPTABLE [add Number to a list, or do a parallel boolean marking 1 for Number] :End
是否支持
、A
、B
和C
以外的字母？您是希望最大化每个组的大小，还是最大化组的数量？问题在于，对于大多数输入，存在多个解决方案。我的问题是：你想要哪一个？您的解决方案也缺少A1。示例中的D
发生了什么？它应该被排除在外吗？如果是，为什么？你需要澄清你想要什么。事实上，没有什么能阻止每个元素都有一个组？@FilipeBorges这对于家庭作业来说似乎太难了（除非它是算法的研究生课程）。如果输出包含A1
非常详细的答案，那就更好了！{A35}，{A36，D37}┌───┬─────┬───┬────┐ │sta│ckove│rfl│owex│ └───┴─────┴───┴────┘(1 0 0 1 0 0 0 0 1 0 0 1 0 0 0) partition (1 1 0 0 1 1 0 1 0 0 0 1 1 0 0)┌─────┬─────────┬─────┬───────┐ │1 1 0│0 1 1 0 1│0 0 0│1 1 0 0│ └─────┴─────────┴─────┴───────┘┌─┬─┬─┬─┐ │2│3│0│2│ └─┴─┴─┴─┘1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 // This would create 1 single, big partition 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 // This would create 15 small partitions// The character part of the argument: 15-length vector of characters: Chars = "A","A","B","D","C","B","A","C","D","C","A","B","A","A","D" // From that, extract the locations of the unique characters: CharsA = 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0 // Where Chars == A CharsB = 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0 // Where Chars == B CharsC = 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0 // Where Chars == C CharsD = 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1 // Where Chars == D // The numeric part of the argument: 15-length vector of numbers: // Btw, is this about lottery... hmmm Nums = 1, 2, 3, 4, 7, 12, 12, 14, 15, 22, 23, 25, 35, 36, 37 :For Number :In [all numbers between 16384 and 32767] Base2 = [2-base of Number] // 20123 would give: 1 0 0 1 1 1 0 1 0 0 1 1 0 1 1 // Test 1: "For each group, it can contains 1 - 4 elements" [using Base2, partition the partition vector Base2 itself; bail out if any partition length exceeds 4] // Test 2: "Difference between element must be <= 3" [using Base2, partition Nums; check differences inside each partition; bail out if bigger than 3 anywhere] // Test 3: "For each group, letter cannot be repeated" [using Base2, partition CharsA, CharsB, CharsC, CharsD (each in turn); count number of ones in each partition; bail out if exceeds 1 anywhere (meaning a character occurs more than once)] // If we still are here, this partition Number is ACCEPTABLE [add Number to a list, or do a parallel boolean marking 1 for Number] :End1 1 0 0 0 1 0 0 0 1 0 0 1 1 0