Kotlin Algorithm Getting Started with Rabbit Number Optimization and Expansion

/* 古典问题：3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少？ 分析：首先我们要明白题目的意思指的是每个月的兔子总对数;假设将兔子分为小中大三种,兔子从出生后三个月后每个月就会生出一对兔子, 那么我们假定第一个月的兔子为小兔子,第二个月为中兔子,第三个月之后就为大兔子,那么第一个月分别有1、0、0,第二个月分别为0、1、0, 第三个月分别为1、0、1,第四个月分别为,1、1、1,第五个月分别为2、1、2,第六个月分别为3、2、3,第七个月分别为5、3、5…… 兔子总数分别为：1、1、2、3、5、8、13……   于是得出了一个规律,从第三个月起,后面的兔子总数都等于前面两个月的兔子总数之和,即为斐波那契数列.*/ class RabbitNumber {

    private var rabbits: Long = 1
    private var lastSecondRabbits: Long = 0
    private var lastRabbits: Long = 0

    /**  * Traverse from the first month to the firstn个月的兔子总数  */  fun forEachMothsToRabbits(moths: Int) {
        println(System.currentTimeMillis())
        for (i in 1..moths) println("第" + i + "个月兔子数为" + getRabbits(i))
        println(System.currentTimeMillis())
    }

    /**  * Get the total number of rabbits for the current month  */  private fun getRabbits(moths: Int): Long {
        if (moths == 1 || moths == 2)
            return rabbits = 1
        else if (moths == 3)
            return rabbits = 2
        else {
            //Initialize the number of rabbits in the last month and the last two months  if (lastRabbits == 0L && lastSecondRabbits == 0L) {
                lastSecondRabbits = getRabbits(moths - 2)
                lastRabbits = getRabbits(moths - 1)
            }
            //Count the number of rabbits returned by this one  rabbits = lastRabbits + lastSecondRabbits  /*Let the number of rabbits in two months equal the number of rabbits in the previous month,Let last month equal the number of rabbits in this month  为了下一次计算（Number of rabbits next month）More efficient and faster to avoid redundant recursion affecting the calculation speed*/  lastSecondRabbits = lastRabbits  lastRabbits = rabbits  return rabbits  }
    }
}