287. Find the Duplicate Number (floyd判圈算法找重复元素)

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

  1. You must not modify the array (assume the array is read only).
  2. You must use only constant, O(1) extra space.
  3. Your runtime complexity should be less than O(n<sup>2</sup>).
  4. There is only one duplicate number in the array, but it could be repeated more than once.

思路:O(n^2)的复杂度暴力即可,说个O(n)复杂度的解法。

需要注意元素范围1..n是个很重要的条件。

使得我们可以把位置和元素映射起来。

可以看成一个迭代函数

由于存在重复重复元素,因此一定存在一个环。

因此可以用floyd判圈算法。

floyd判圈算法_维基百科

这算法以前遇到过sgu455 解题报告,所以不算很陌生…

但是从有重复元素没有联想到会有环orz,我好菜啊…

 

sgu 455. Sequence analysis (floyd 判圈算法,O(1)空间复杂度求循环节)

455. Sequence analysis

Time limit per test: 1 second(s)
Memory limit: 4096 kilobytes
input: standard
output: standard

Due to the slow ‘mod’ and ‘div’ operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not guarantee that Delphi solution exists. 

You are given a sequence of signed 64-bit integers defined as follows:

  • x0 = 1,
  • ,

where

mod

is a remainder operator. All arithmetic operations are evaluated without overflow checking. Use standard “remainder” operator for programming languages (it differs from the mathematical version; for example  in programming, while  in mathematics).

Let’s call a sequence element xp repeatable if it occurs later in the sequence — meaning that there exists such qq > p, that xq = xp. The first repeatable element M of the sequence is such an element xmthat xm is repeatable, and none of the xp where p < m are repeatable.

Given AB and C, your task is to find the index of the second occurence of the first repeatable element M in the sequence if the index is less or equal to 2 · 106. Per definition, the first element of the sequence has index 0.

Input

The only line of input contains three signed 64-bit integers: AB and C (B > 0, C > 0).

Output

Print a single integer  — the index of the second occurence of the first repeatable member if it is less or equal to 2 · 106. Print -1 if the index is more than 2 · 106.

Example(s)

 

 

 

Note

In the first sample test the sequence starts with the following numbers: 1, 3, 7, 6, 3, 7. The first repeatable element is 3. The second occurence of 3 has index 4.

In the second sample test the sequence starts with the following numbers: 1, 2305843009213693951, -4611686018427387903, 6917529027641081855, 0, 0, 0. The first repeatable element is 0. The second occurence of 0 has index 5.

In the third sample test the sequence starts with the following numbers: 1, -2, 4, -3, 1, -2, 4. The first repeatable element is 1. The second occurence of 1 has index 4.

 

比赛的时候逗了,往看空间限制了….

直接开了个set判重。。。显然MLE 了。。。

然后这道题的正解是 floyd判圈算法(也叫龟兔算法?)

http://www.cnblogs.com/oyking/p/4286916.html  这份题解讲得很详细