问题描述：
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:
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
在一个数组中找到重复的那个。
由于题目限制了只能使用O(1)的空间，也不能修改参数数组。
因此不能排序，也不能创造一个标记数组来一次遍历进行'填空'；并且，由于重复元素可以出现多次，也无法使用等差数列公式求和=>s，再用数组和-s的方式。
搜到了一种解法，比较巧妙，思路类似于那个题目：在一个链表中找到环，并找出那个环的位置。
1. 使用了快慢指针
2. 在快慢指针相遇的位置开始，慢指针与另一个从起始位置出发的指针每次走一步，直到它们相遇，就是那个重复节点发生的地方。
以下实现参考了链接：
http://bookshadow.com/weblog/2015/09/28/leetcode-find-duplicate-number/
实现代码：public class Solution {
public int FindDuplicate(int[] nums)
{
// define slow and fast , fast each time move 2 times
var slow = 0;
var fast = 0;
while(true)
{
slow = nums[slow];
fast = nums[nums[fast]];
if(slow == fast){
break;
}
}
// now, lets create another pointer 'finder' from the start position.
// slow will stay where it is.
// finder and slow each time move one step. next time this meet is where duplicate happens
var finder = 0;
while(true)
{
slow = nums[slow];
finder = nums[finder];
if (slow == finder){
return slow;
}
}
}
}