本文探讨了在有限内存环境下，通过使用快慢指针法解决循环调用函数产生的环形序列问题，包括计算环长、循环起点及避免冗余存储的方法。
//Q: Given a function that takes in an integer input and gives out an integer:
//int func(int input);
//This function is deterministic. It always gives the same output value for a given input. Say func(8) = 7 whenever it is called.
//Now this function is called multiple times in a loop. This would generate a stream of integers, what is the length of this loop ? give me the count of how many numbers are repeated and how many
//numbers are not repeated. He said there is very limited memory: you can store only a certain number of integers at a time, but not all of them.
//int func(int input);
//This function is deterministic. It always gives the same output value for a given input. Say func(8) = 7 whenever it is called.
//Now this function is called multiple times in a loop. This would generate a stream of integers, what is the length of this loop ? give me the count of how many numbers are repeated and how many
//numbers are not repeated. He said there is very limited memory: you can store only a certain number of integers at a time, but not all of them.
就是说有一个函数 int func(int input), 给一个输入值，返回一个输出，新的输出会成为下一个的输入。在函数的调用过程中总会进入一个loop，判断对于一个给定的输入值，这样循环调用产生的loop的长度
看起来非常像那个经典的链表找环问题，那题用快慢两个指针，我觉得这题也能用类似的办法。用两个变量，初始a=b=input，每次迭代中a=f(a),b=f(f(b))，如果函数保证总会进loop，那a和b总会相等。就能得到loop长度的一个整倍数，然后能找到loop起点，然后就能O(1)空间找到长度了
扫一扫
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
