思路很直接，统计数组中每一个值的频率，创建一个哈希表用空间换时间，优化查询效率
def solution(array):
# Edit your code here
n=len(array)
hash_dict={}
for value in array:
hash_dict[value]=hash_dict.get(value,0)+1
if hash_dict[value]/n>0.5:
return value
return 0
if __name__ == "__main__":
# Add your test cases here
print(solution([1, 3, 8, 2, 3, 1, 3, 3, 3]) == 3)
