LeetCode每日一题——T1. 两数之和（易）：三种解法：暴力循环、两遍哈希表、一遍哈希表

要点：enumerate()函数、len()函数、创建空字典：_dict = {}、哈希表用法：_dict.get(值)

# 1、暴力解法：

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i, m in enumerate(nums):
for j in range(i, len(nums)):
if (nums[j] == target - m):
return [i, j]

# enumerate: 对于一个可迭代的（iterable）/可遍历的对象（如列表、字符串）, 将其组成一个索引序列，利用它可以同时获得索引和值;如上，i是索引，m是值

# 2、两遍哈希表：

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
# 定义空字典，用以模拟哈希表
_dict = {}
for i, m  in enumerate(nums):
_dict[m] = i

for i, m in enumerate(nums):
j = _dict.get(target - m)
if j is not None and j != i:
return[i, j]

# 3、一遍哈希表：

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
_dict = {}
for i, m  in enumerate(nums):
if _dict.get(target - m) is not None:
return [_dict.get(target - m), i]
_dict[m] = i