LeetCode/1. 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 1:

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

Constraints:

- 2 <= nums.length <= 10^4
- 10^9 <= nums[i] <= 10^9
- 10^9 <= target <= 10^9
- Only one valid answer exists.

Follow-up:
Can you come up with an algorithm that is less than O(n2) time complexity?
Approach 2: Two-pass Hash Table
Complexity Analysis

Time complexity: O(n)

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        딕셔너리 = {}
        for i in range(len(nums)) :
            딕셔너리[nums[i]] = i
        for i in range(len(nums)) :
            찾는값 = target - nums[i]
            if 찾는값 in 딕셔너리 and 딕셔너리[찾는값] != i
                return [i, 딕셔너리[찾는값]]

What is the best way to maintain a mapping of each element in the array to its index?
A hash table.
ハッシュ表.勉強するべきでしょう.
In the first iteration, we add each element's value as a key and its index as a value to the hash table
最初の繰り返し文では、
To the hashテーブルは,各データをキー値とし,インデックスを値とする.
Then, in the second iteration, we check if each element's complement ( target - nums[i] ) exists in the hash table.
2番目の繰り返し文では、
検索する数値が存在するかどうかを確認します.
インデックス番号を比較して、異なるデータを確認
If it does exist, we return current element's index and its complement's index. Beware that the complement must not be nums[i] itself!
存在する場合、
現在回転しているやつと検索する数字のインデックス番号を返します.
Approach 3: One-pass Hash Table
Complexity Analysis

Time complexity: O(n)

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hashmap = {}
        for i in range(len(nums)) :
            complement = target - nums[i]
            if complement in hashmap :
                return [i, hashmap[complement]]
            hashmap[nums[i]] = i

きれい...
れんが1枚
range()
整数シーケンス

の

を生成する.

for i in range(4):
    print(i) #0 1 2 3

for i in range(0, 4):
    print(i) # 0 1 2 3

for i in range(2, 6):
    print(i) # 2 3 4 5

len()
anobject

における物品数

list = ['a', 'b', 'c', 'd']

printme = len(list)

print(printme) # 4

range(len(リスト))

リストのアイテム数のrange!!

list = ['a', 'b', 'c', 'd']

for i in list :
    print(i) # a b c d 
for i in range(len(list)) :
    print(i) # 0 1 2 3

キーの検索

バイナリファイルにキー値(真/偽)があるかどうかを確認する

#
찾는key in 딕셔너리
#
if 찾는key in 딕셔너리 : print('있네염')
else : print('없네여')

小毛の子.
1日に1つの道しるべ.
0.2個も解けない