Find the Duplicate Number

https://leetcode.cn/problems/find-the-duplicate-number/submissions/

方法1：pos记录无重复区间的下一个位置 （TLE）

    def findDuplicate(self, nums: List[int]) -> int:
        pos = 0
        for i in range(len(nums)):
            if nums[i] not in nums[:pos]:
                nums[pos], nums[i] = nums[pos], nums[i]
                pos += 1
            else:
                return nums[i]

方法2：二分查找
left, right = 1, len(nums) # 重复元素一定在1到n之间
1.假设mid是重复元素；统计数组中小于等于mid的元素个数cnt
2.如果cnt > mid，那么重复元素一定位于 [left...mid]
3.如果cnt < mid, 那么重复元素一定位于 [mid+1 ...right]

 def findDuplicate(self, nums: List[int]) -> int:
        # 1 2 2
        # mid = 2
        # cnt = 3  > mid ---> right = mid

        left, right = 1, len(nums) 
        while left < right:
            mid = (left + right) >> 1
            cnt = 0

            for num in nums:
                if num <= mid:
                    cnt += 1
            
            if cnt <= mid:
                left = mid + 1
            else:
                right = mid 
            
        return left

方法3：快慢指针找环的入口

    def findDuplicate(self, nums: List[int]) -> int:
        # https://leetcode.cn/problems/find-the-duplicate-number/solution/287xun-zhao-zhong-fu-shu-by-kirsche/
        slow, fast = nums[0], nums[nums[0]]
        while slow != fast:
            slow = nums[slow]
            fast = nums[nums[fast]]
        
        slow = 0
        while slow != fast:
            slow = nums[slow]
            fast = nums[fast]
        
        return slow