https://leetcode.com/problems/find-the-duplicate-number/?tab=Description
没想到的是，居然重复复用了链表环形追逐的思路。非常的巧妙。而且不需要修改数组里面的数字，不需要修改原数组。
https://discuss.leetcode.com/topic/25913/my-easy-understood-solution-with-o-n-time-and-o-1-space-without-modifying-the-array-with-clear-explanation/2
int findDuplicate3(vector<int>& nums) { if (nums.size() > 1) { int slow = nums[0]; int fast = nums[nums[0]]; while (slow != fast) { slow = nums[slow]; fast = nums[nums[fast]]; } fast = 0; while (fast != slow) { fast = nums[fast]; slow = nums[slow]; } return slow; } return -1; }
原文：http://www.cnblogs.com/charlesblc/p/6445526.html