Java：算法 - 开局一对兔，两月就成熟

题目：古典问题：有一对兔子，从出生后第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，问每个月的兔子对数为多少？
程序分析：哇是斐波那契数列1,1,2,3,5,8,13,21… 用递归fn = fn-1 +fn-2

吐槽：题干别读成小兔子长到第三个月后（的第四个月）又生了一对兔子，这样递归会变成fn = fn-2+fn-3+fn-4…，小兔子是2周月就成熟！

递归解法：从第三月开始，总数为前2个月的和：

	public static int countPairsOfRabbits(int month) throws Exception{
		if(month == 1 || month == 2){
			return 1;
		}else if(month <=0){
			throw new Exception("invalid argument of month");
		}else{
			return countPairsOfRabbits(month-1)+countPairsOfRabbits(month-2);
		}
	}

如果要问每个月小兔子和大兔子的对数怎么办？

对数： 小兔+大兔=兔子
初始1对小兔，花2个月变成大兔子，然后大兔子生同样数量的小兔。

可以把小兔放进一个长度2的数组，模拟队列，出列合并大兔生仔~

	public static void countRabbitAndBunny(int month) throws Exception{

		if(month == 1 || month == 2){
			System.out.println("Pairs: Sum: 1 || Rabbit: 0 || Bunny: 1");
		}else if(month <=0){
			throw new Exception("invalid argument of month");
		}else{
			//where we found our bunnies
			int[] bunny = new int[2];
			
			/*
			 * the second month status
			 */
			bunny[1] = 1; //bunny of month 2 in the array  
			int pairOfBunny = 1;
			int pairOfRabbit = 0;	
			int sum = pairOfBunny + pairOfRabbit;
			
			//count the "i"th month status
			for(int i=3;i<=month;i++){
				//bunnies became rabbits
				pairOfRabbit = pairOfRabbit + bunny[1];
				pairOfBunny = pairOfBunny - bunny[1];
				
				//some bunnies grew up
				bunny[1] = bunny[0];
				//some bunnies were just born
				bunny[0] = pairOfRabbit;
				pairOfBunny = pairOfBunny + bunny[0];
				
				//count all rabbits alive
				sum = pairOfBunny + pairOfRabbit;
			}
			
			//we now have the result
			System.out.println("Pairs: Sum: " + sum + "|| Rabbit:  "
			+ pairOfRabbit + " || Bunny: "+pairOfBunny);
		}
	}