import pytest


def twoSum(nums, target):
    lens = len(nums)
    j=-1
    for i in range(lens):
        if (target - nums[i]) in nums:
            # 如果num2=num1,且nums中只出现了一次，说明找到是num1本身
            if (nums.count(target - nums[i]) == 1) & (target - nums[i] == nums[i]):
                continue
            else:
                # index(x,i+1)是从num1后的序列找num2
                j = nums.index(target - nums[i], i+1)
                break
    if j > 0:
        return [i, j]
    else:
        return []

#
# def twoSum(nums, target):
#     hashmap={}

      #enumerate() 函数用于将一个可遍历的数据对象(如列表、元组或字符串)组合为一个索引序列，同时列出数据和数据下标
#     for ind,num in enumerate(nums):
#         hashmap[num] = ind
#     for i,num in enumerate(nums):
#         j = hashmap.get(target - num)
#         if j is not None and i!=j:
#             return [i,j]


def test_case_1():
    res1 = twoSum([2, 3, 1, 5, 8], 6)
    expect1 = [2, 3]
    assert expect1 == res1

@pytest.mark.parametrize(
    "arr, target, expect",
    [
        ([-2, 3, 1, 5, 8], -1, [0, 2]),
        ([2, 3, 1, 5, 8], 6, [2, 3])
    ]
)
def test_cases(arr, target, expect):
    assert twoSum(arr, target) == expect
#
# def test_case_2():
#     res2 = twoSum([-2, 3, 1, 5, 8], -1)
#     expect2 = [0, 2]
#     assert expect2 == res2
#
#     res3 = twoSum([2, 1, 1, 5, 8], 2)
#     expect3 = [1, 2]
#     assert expect3 == res3
#
#     res4 = twoSum([2, -1, 1, 5, 8], 0)
#     expect4 = [1, 2]
#     assert expect4 == res4
#
#     res5 = twoSum([2, 0, 0, 1, -1], 0)
#     expect5 = [1, 2]
#     assert expect5 == res5
#
#
#