Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: 
Because nums[0] + nums[1] == 9, we return [0, 1]

My solution

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        length = len(nums) - 1
        index = 0
        answer_list = []

        for num_first in nums:
            while length > -1:
                if (num_first + nums[length] == target) and (index != length):
                    answer_list.append(index)
                    answer_list.append(length)
                    break

                length -= 1
            if len(answer_list) != 0:
                break
            else:
                index += 1
                length = len(nums) - 1

        return answer_list

Best solution

def twoSum(self, nums, target):
    hash_map = {}
    for i in range(len(nums)):
        if target - nums[i] in hash_map:
            return [hash_map[target - nums[i]], i]
        else:
            hash_map[nums[i]] = i
# 这个算法使用了哈希表来存储已经遍历过的数字及其索引，
# 以便快速查找目标值减去当前数字的差值是否在数组中。
# 时间复杂度为 O(n)。

在 Python 中，hash_map 通常指的是字典（dictionary）类型。字典是一种可变容器模型，可以存储任意类型的键值对。字典中的键必须是唯一的，但值则不必。

# 创建一个空字典
my_dict = {}

# 向字典中添加键值对
my_dict['key1'] = 'value1'
my_dict['key2'] = 'value2'

# 访问字典中的值
print(my_dict['key1']) # 输出 'value1'

# 删除字典中的键值对
del my_dict['key1']

# 检查键是否在字典中
if 'key1' in my_dict:
    print('Key found')
else:
    print('Key not found') # 输出 'Key not found'

您可以使用 in 关键字来检查某个键是否在字典中。您还可以使用 del 关键字来删除字典中的某个键值对。