LeetCode #287 - Find the Duplicate Number

题目描述：

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:

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(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

给定n+1个数，取值范围为[1,n]，其中一个数出现两次，其他所有数都只出现一次，求这个数。

由于数组取值范围都特殊性，可以令元素都取值为下一个元素的指针，即元素的值为下一个元素的下标，例如nums=[2,4,5,3,5,1]，那么可看作链表2->5->1->4->5，可见重复的数为链表的环的起点，所以可以采用求链表的环的起点的方法。

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int slow=0;
        int fast=0;
        while(true)
        {
            slow=nums[slow];
            fast=nums[nums[fast]];
            if(nums[slow]==nums[fast]) break;
        }
        int temp=0;
        while(nums[temp]!=nums[slow])
        {
            slow=nums[slow];
            temp=nums[temp];
        }
        return nums[slow];
    }
};