287. Find the Duplicate Number

Description：
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.
Difficulty：Medium

Input: [1,3,4,2,2]
Output: 2
Input: [3,1,3,4,2]
Output: 3

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 $O($

方法1：排序、Hash、取負標記，不符合題意

思路：
很容易聯想到題目：448.Find All Numbers Disappeared in an Array
利用改變元素正負號和取絕對值的方法，標記是否出現過，如果出現過，直接返回index，但是不符合題意，改變了原陣列。

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
		for (int i = 0; i < nums.size(); i++) {
			int index = abs(nums[i]) - 1;
			if (nums[index] < 0) 
				return index + 1;
			else
				nums[index] = -nums[index];
		}
    }
};

方法2：快慢指標

• Time complexity : $O\left ( n \right )$
• Space complexity : $O\left ( 1 \right )$
  思路：
  嘗試連線成環，肯定能夠連線成以重複元素作為入口的環。
  接下來問題就變成了找環的入口，經典公式，第一次快慢指標相遇時a+x=(a+b+x)/2 => a+x=b，然後快指標歸0,並且變成慢指標速度，快慢指標同時走，第二次相遇便是入口。
  關鍵點：

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