class Solution:
    def duplicate(self , numbers ):
        # write code here
#         dic = {}                 # 方法1:使用dic字典,如果存在则返回,不存在则标记,空间复杂度o(n)
#         if not numbers:
#             return -1
#         for i in range(len(numbers)):
#             if numbers[i] in dic:
#                 return numbers[i]
#             else:
#                 dic[numbers[i]] =1

         if not numbers:          # 方法二:空间复杂度o(1),遍历原数组,将原数组的值作为数组的索引,如果大于0表示第一次访问,然后取反,如果小于0表示已经访问过了。
             return -1
         for i in range(len(numbers)):
            if numbers[abs(numbers[i])]<0:
                return abs(numbers[i])
            else:
                numbers[abs(numbers[i])] = -numbers[abs(numbers[i])]