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## Problem

Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one duplicate number in nums, return this duplicate number.

Follow-ups:

How can we prove that at least one duplicate number must exist in nums? Can you solve the problem without modifying the array nums? Can you solve the problem using only constant, O(1) extra space? Can you solve the problem with runtime complexity less than O(n2)?

Example 1:

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


Example 2:

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


Example 3:

Input: nums = [1,1]
Output: 1


Example 4:

Input: nums = [1,1,2]
Output: 1


## Code

class Solution {
public int findDuplicate(int[] nums) {
int slow = 0;
int fast = 0;

// 找到在环上的某一点相遇
while(true){
slow = nums[slow];
fast = nums[nums[fast]];
if (slow == fast) {
break;
}
}

// 找到进入环的点
fast = 0;
while(true){
slow = nums[slow];
fast = nums[fast];
if (slow == fast) {
break;
}
}

return slow;
}
}

class Solution:
def findDuplicate(self, nums: List[int]) -> int:
slow = nums[0]
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

func findDuplicate(nums []int) int {
slow := 0
fast := 0

for {
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast {
break
}
}

fast = 0
for {
slow = nums[slow]
fast = nums[fast]
if slow == fast {
break
}
}

return slow
}