leetcode/array/two_sum.py

import pytest


# 方法1
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 []

# 方法2
# 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]


@pytest.mark.parametrize(
    "arr, target, expect",
    [
        ([-2, 3, 1, 5], -1, [0, 2]),
        ([2, 3, 1, 5], 6, [2, 3]),
        ([2, 1, 1, 5, 8], 2, [1, 2]),
        ([2, -1, 1, 5, 8], 0, [1, 2]),
        ([2, 0, 0, 1, -1], 0, [1, 2])
    ]
)
def test_cases(arr, target, expect):
    assert twoSum(arr, target) == expect