最简单的第一题，想要有好的时间空间复杂度也是挺难得

    1. #暴力解1  用了两层for循环
    2. def twoSum(self, nums: list[int], target: int) -> list[int]:
    3.     for i in range(len(nums)-1):
    4.         for j in range(i+1, len(nums)):  #有个问题，最后两个元素不会被遍历
    5.             if nums[i]+nums[j] == target:
    6.                 return [i, j]
    9. #暴力解2 使用了列表内置的count 和 index
    10. class Solution:
    11.     def twoSum(self, nums: List[int], target: int) -> List[int]:
    12.         for i in range(len(nums)-1):
    13.                 if target-nums[i] in nums:
    14.                     if nums.count(target-nums[i]) == 1 and 2*nums[i] == target: #说明只有一个
    15.                         continue
    16.                     j = nums.index(target-nums[i], i+1, len(nums))
    17.                     return [i, j]
    18.         return []
    20. #暴力解3   优化了方法2   优化了索引
    21. class Solution:
    22.     def twoSum(self, nums: List[int], target: int) -> List[int]:
    23.         for i in range(len(nums)-1):
    24.                 if target-nums[i] in nums[i+1:]:
    25.                     j = nums.index(target-nums[i], i+1, len(nums))
    26.                     return [i, j]
    27.         return []
    31. #解4  采用了字典模仿哈希
    32. class Solution:
    33.     def twoSum(self, nums: List[int], target: int) -> List[int]:
    34.         dic = {}
    35.         for i,n in enumerate(nums):
    36.             cp = target - n
    37.             if cp in dic:
    38.                 return [dic[cp], i]
    39.             else:
    40.                 dic[n] = i  #真的绝，用值作为键，用索引作为值，直接访问