Leetcode sum of two numbers

Given an array of integers nums  And an integer target value target, Please find... In the array And is the target value target   the   Two   Integers , And return their array subscripts .

You can assume that each input corresponds to only one answer . however , The same element in the array cannot be repeated in the answer .

You can return the answers in any order .

My answer

Example 1：

Input ：nums = [2,7,11,15], target = 9
Output ：[0,1]
explain ： because nums[0] + nums[1] == 9 , return [0, 1] .
Example 2：

Input ：nums = [3,2,4], target = 6
Output ：[1,2]
Example 3：

Input ：nums = [3,3], target = 6
Output ：[0,1]
 

Tips ：

2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
There will only be one valid answer
Advanced ： You can come up with a time complexity less than O(n2) The algorithm of ？

My answer ：

class Solution:
    def twoSum(self,nums,target):
       list1=[]
        
       for x in nums:
           for y in nums:
               if (x+y==target):
                  list1.append(nums.index(x))
                  list1.append(nums.index(y))
                  return list1
                
        

       

The function I wrote , By traversing the list directly , Find the number first , Then enter the value index. Time complexity O(n Fang )

Input example 1 , Get the results you want . But it can be seen that , Yes, there is bug Of . If you enter an instance 2, Will be output [0,0] But this is not the desired result . so , When it comes to testing , You need to know the internal logic of the code , Otherwise, the selected test case data , The test passed . But the actual code bug Not tested .

 

  The following solution , It won't exist index Repetitive questions . But the time complexity is still On Fang

class Solution:
    def twoSum(self,nums,target):
        list1=[]
        for i in range(len(nums)-1):
            for j in range(i+1,len(nums)):
              if (nums[i]+nums[j]==target):
                    return[i,j]
        return []
        

                
        

       

  Solution 2 of the answer ：

Hashtable

 

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hashtable = dict()
        for i, num in enumerate(nums):
            if target - num in hashtable:
                return [hashtable[target - num], i]
            hashtable[nums[i]] = i
        return []

 Python There is no hash in , A dictionary is equivalent to a hash .

But I don't understand it here ,hashtable It's an empty dictionary , How to follow enumerate(nums) Related ？

Turn lists into dictionaries and lists . 

  I can't understand this writing , Put it down first , See if you can understand it later .