전체 글 (306)
2022-01-28 04:57:24
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2022-01-28 04:46:46
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Consecutive Numbers

 

 문제 설명

 

Table: Logs

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| num         | varchar |
+-------------+---------+
id is the primary key for this table.

 

Write an SQL query to find all numbers that appear at least three times consecutively.

Return the result table in any order.

The query result format is in the following example.

 

 입출력 예

 

Example 1:

Input: 
Logs table:
+----+-----+
| id | num |
+----+-----+
| 1  | 1   |
| 2  | 1   |
| 3  | 1   |
| 4  | 2   |
| 5  | 1   |
| 6  | 2   |
| 7  | 2   |
+----+-----+
Output: 
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+
Explanation: 1 is the only number that appears consecutively for at least three times.

 

 Oracle Query

 

SELECT DISTINCT 
	l1.Num As ConsecutiveNums
FROM Logs l1
JOIN Logs l2 ON l1.Id = l2.Id-1
JOIN Logs l3 ON l2.Id = l3.Id-1
WHERE l1.Num = l2.Num
	AND l2.Num = l3.Num;

* 참고 링크 : https://leetcode.com/problems/consecutive-numbers/discuss/185886/Self-Join-Twice-(681-ms-faster-than-100.00-of-Oracle-online-submissions-for-Consecutive-Numbers.)

 

 출처

 

https://leetcode.com/problems/consecutive-numbers/

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2022-01-28 04:23:15
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Valid Triangle Number

 

 문제 설명

 

Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

 

 제한 사항

 

  • 1 <= nums.length <= 1000
  • 0 <= nums[i] <= 1000

 

 입출력 예

 

Example 1:

Input: nums = [2,2,3,4]
Output: 3
Explanation: Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Example 2:

Input: nums = [4,2,3,4]
Output: 4

 

 Python 코드

 

Python code 

class Solution(object):
    def triangleNumber(self, nums):

        nums, count, n = sorted(nums, reverse=1), 0, len(nums)
        for i in range(n):
            j, k = i + 1, n - 1
            while j < k:
                if nums[j] + nums[k] > nums[i]:
                    count += k - j
                    j += 1
                else:
                    k -= 1
        return count

* 참고 링크 : https://leetcode.com/problems/valid-triangle-number/discuss/104177/O(N2)-solution-for-C%2B%2B-and-Python

 

 C++ 코드

 

C ++ code

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        if(nums.size() < 3) return 0;
        sort(nums.begin(), nums.end(), greater<int>());
        int res = 0;
        int n = nums.size();
        for(int a = 0; a < n-2; ++a){
            int b = a+1;
            int c = n-1;
            while(b < c){
                if(nums[b] + nums[c] > nums[a]){
                    res = res + c - b;
                    ++b;
                }
                else
                    --c;
            }
        }
        return res;
    }
};

* 참고 링크 : https://www.programmerall.com/article/1415760413/

 

 출처

 

https://leetcode.com/problems/valid-triangle-number/

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2022-01-28 03:38:43
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Find Peak Element

 

 문제 설명

 

A peak element is an element that is strictly greater than its neighbors.

Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞.

You must write an algorithm that runs in O(log n) time.

 

 제한 사항

 

  • 1 <= nums.length <= 1000
  • -231 <= nums[i] <= 231 - 1
  • nums[i] != nums[i + 1] for all valid i.

 

 입출력 예

 

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

 

 Python 코드

 

Python code 

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left = 0
        right = len(nums)-1
				# left가 커지고 right가 작아지는 경우 left와 right가 같아질 수 있다.
        # 그 때 원한느 값을 찾은 경우이므로 left와 right가 같아지면 반복문 종료
        while left < right:
            mid = (left + right) // 2      # 가운데 지점
            if nums[mid] < nums[mid+1]:    # 현재 값이 오른쪽 값보다 작다면
                left = mid+1
            else:                          # 현재 값이 오른쪽 값보다 크다면
                right = mid
        return left                        # 반복문이 종료되면 left가 정답

* 참고 링크 : https://velog.io/@hrpp1300/LeetCode-162.-Find-Peak-Element

 

 C++ 코드

 

C ++ code

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > nums[mid + 1]) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        return left; 
    }
};

* 참고 링크 : https://myeongcs.tistory.com/80

 

 출처

 

https://leetcode.com/problems/find-peak-element/submissions/

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2022-01-28 03:12:48
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Same Tree

 

 문제 설명

 

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.

You call a pre-defined API int guess(int num), which returns three possible results:

  • -1: Your guess is higher than the number I picked (i.e. num > pick).
  • 1: Your guess is lower than the number I picked (i.e. num < pick).
  • 0: your guess is equal to the number I picked (i.e. num == pick).

Return the number that I picked.

 

 제한 사항

 

  • 1 <= n <= 231 - 1
  • 1 <= pick <= n

 

 입출력 예

 

Example 1:

Input: n = 10, pick = 6
Output: 6

Example 2:

Input: n = 1, pick = 1
Output: 1

Example 3:

Input: n = 2, pick = 1
Output: 1

 

 

 Python 코드

 

Python code 

# The guess API is already defined for you.
# @param num, your guess
# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
# def guess(num: int) -> int:

class Solution: 
    def guessNumber(self, n: int) -> int: 
        left, right = 1, n 
        
        while True: 
            middle_num = int((left + right) / 2) 
            if guess(middle_num) == 1: 
                left = middle_num + 1 
            elif guess(middle_num) == -1: 
                right = middle_num - 1 
            else: 
                return middle_num

* 참고 링크 : https://somjang.tistory.com/entry/leetCode-374-Guess-Number-Higher-or-Lower-Python

 

 C++ 코드

 

C ++ code

//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Guess Number Higher or Lower.
//Memory Usage: 7.3 MB, less than 100.00% of C++ online submissions for Guess Number Higher or Lower.

// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);

class Solution {
public:
    int guessNumber(int n) {
        int left = 1, right = n, cur = left+(right-left)/2;

        while(true){
            switch(guess(cur)){
                case -1:
                    right = cur-1;
                    cur = left+(right-left)/2;
                    break;
                case 1:
                    left = cur+1;
                    cur = left+(right-left)/2;
                    break;
                case 0:
                    return cur;
            }
        }
        return 0;
    }
};

* 참고 링크 : https://github.com/keineahnung2345/leetcode-cpp-practices/blob/master/374.%20Guess%20Number%20Higher%20or%20Lower.cpp

 

 출처

 

https://leetcode.com/problems/guess-number-higher-or-lower/

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2022-01-26 08:16:15
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DOM(Document Object Model) ?

DOM(Document Object Model)은 문서 객체 모델로 XML, HTML 문서의 각 항목을 계층으로 표현하여 생성, 변형, 삭제할 수 있도록 돕는 인터페이스이다. W3C의 표준이다.

 

 DOM TREE

 

<html>
	<head>
    	<title>Hello</title>
    </head>
    <body>
    	<p>aaa</p>
        <p>bbb</p>
        <span id="x">ccc</span>
        <span id="y">ddd</span>
        <a href="https://www.naver.com">네이버</a>
    </body>
</html>

 출처

 

DOM 정의 -  https://ko.wikipedia.org/wiki/%EB%AC%B8%EC%84%9C_%EA%B0%9D%EC%B2%B4_%EB%AA%A8%EB%8D%B8

https://post.naver.com/viewer/postView.nhn?volumeNo=25195764 

 

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