Kotlin算法入门求回文数算法优化一

class PalindromeNumber {

    /**  * 创建map集合存储数据然后对数据进行判断抛出  */  private val palindromeData = HashMap<Int, ArrayList<String>>()

    /**  * 讲数字直接转为字符串然后通过字符串中间截取用两个首尾进行对比得出  */  fun isPalindromeNumber(number: Long): Boolean {
        //个位直接返回 小于100的回文数都是可以整除11 整除10的数绝对不是回文数  if (number < 10) return true  if (number < 100) return number!! % 11 == 0L
        if (number!! % 10 == 0L) return false  val numbers = number.toString() + ""  /*三位数回文数例如：121 131 151等等都是前后相同则直接判断是否前后相同相同直接返回  * 然后符合规则的数字放入一个key为5的map里面一个List集合中去  * 当number为5为数字的时候判断第一个数根最后一个数是否相同  * 相同则去除首尾取中间进行对比  * 是否map中key为5的集合中是否含有余下的这三位数有则返回true反之flase  * 其中返回前如果true则放入一个key为7的List中去进行递归字典处理  * 偶数位的回文数与基数位的回文数处理类似所以不在讲解*/  if (number > 100 && number <= 999) {
            if (number / 100 == number % 10) {
                (palindromeData as java.util.Map<Int, ArrayList<String>>).computeIfAbsent(5) { k -> ArrayList() }
                palindromeData[5]!!.add(numbers)
                return true  } else  return false  }
        if (number >= 1001 && number <= 9999) {
            if (number / 1000 == number % 10 && number / 100 % 10 == number % 100 / 10) {
                (palindromeData as java.util.Map<Int, ArrayList<String>>).computeIfAbsent(6) { k -> ArrayList() }
                palindromeData[6]!!.add(numbers)
                return true  }
            return false  }
        val firstNumber = numbers.substring(0, 1)
        val lastNumber = numbers.substring(numbers.length - 1, numbers.length)
        if (firstNumber != lastNumber) return false  val length = numbers.length  val subString = numbers.substring(1, length - 1)
        return if (Integer.parseInt(subString) == 0) true else ismapData(numbers, length, number, subString)

    }

    private fun baseIsPalindromeNumber(numberFrist: String, numbersLast: String, numbersLast2: String): Boolean {
        var numbersLast2 = numbersLast2
        var i = 0
        while (i < numbersLast.length) {
            val index = numbersLast.length - i - 1
            if (i == 0) {
                /*为了避免进行大数据类型遍历的加上第一位数的判断如果不相同直接返回以此减少每一次需要进行数据逆转的次数  数据长度的记为n 则减少n/2-1次的判断次数  */  numbersLast2 = numbersLast2 + numbersLast.substring(index)
                if (numberFrist.substring(0, 1) != numbersLast2) return false  } else {
                numbersLast2 = numbersLast2 + numbersLast.substring(index, index + 1)
                if (index == 2 || index == 5 || index == 8) {
                    if (numberFrist.substring(0, numbersLast2.length) != numbersLast2) return false  }
            }
            ++i
        }
        return numberFrist == numbersLast2
    }

    private fun olderIsPalindromeNumber(number: Long): Boolean {
        //个位直接返回 小于100的回文数都是可以整除11 整除10的数绝对不是回文数  // System.out.println(number+"num");  if (number < 10) return true  if (number < 100) return number!! % 11 == 0L
        if (number!! % 10 == 0L) return false  val numbers = number.toString() + ""  // System.out.println(numbers);  /*System.out.println("number = [" + numbers.length() + "]");*/  val halfIndex = numbers.length / 2
        val numberFrist = numbers.substring(0, halfIndex - 1)
        var numbersLast = ""  val numbersLast2 = ""  //对最后字符串截取需要注意的是字符串长度是否为偶数  if (numbers.length / 2 == 0) {
            numbersLast = numbers.substring(halfIndex - 1)
        } else {
            numbersLast = numbers.substring(halfIndex + 1)
        }

        // System.out.println(numberFrist+ "ssss "+numbersLast);  return baseIsPalindromeNumber(numberFrist, numbersLast, numbersLast2)
    }

    private fun ismapData(numbers: String, length: Int, number: Long, subString: String): Boolean {
        if (palindromeData[length] != null) {
            // System.out.println("notnull sub:"+numbers.substring(1,length-1));  if (palindromeData[length]!!.contains(subString)) {
                (palindromeData as java.util.Map<Int, ArrayList<String>>).computeIfAbsent(length + 2) { k -> ArrayList() }
                palindromeData[length + 2]!!.add(numbers)
                return true  } else {
                val next = Math.pow(10.0, (length - 1).toDouble()).toLong()
                val last = Math.pow(10.0, (length - 2).toDouble()).toLong()
                if (number > next + last + 11) return false  if (subIsPalindromeNumber(numbers)) {
                    (palindromeData as java.util.Map<Int, ArrayList<String>>).computeIfAbsent(length + 2) { k -> ArrayList() }
                    palindromeData[length + 2]!!.add(numbers)
                    palindromeData[length]!!.add(subString)
                    return true  }
                return false  }
        } else {
            if (subIsPalindromeNumber(numbers)) {
                (palindromeData as java.util.Map<Int, ArrayList<String>>).computeIfAbsent(length + 2) { k -> ArrayList() }
                palindromeData[length + 2]!!.add(numbers)
                return true  } else  return false  }
    }

    private fun subIsPalindromeNumber(numbers: String): Boolean {
        val halfIndex = numbers.length / 2
        val numberFrist = numbers.substring(0, halfIndex - 1)
        var numbersLast = ""  val numbersLast2 = ""  //对最后字符串截取需要注意的是字符串长度是否为偶数  if (numbers.length / 2 == 0) {
            numbersLast = numbers.substring(halfIndex)
        } else {
            numbersLast = numbers.substring(halfIndex)
        }
        return baseIsPalindromeNumber(numberFrist, numbersLast, numbersLast2)
    }
}