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1 月 2021

LeetCode: 16-3Sum Closest 解題紀錄

Last Updated on 2021-07-28 by Clay


題目

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Constraints:

  • 3 <= nums.length <= 10^3
  • -10^3 <= nums[i] <= 10^3
  • -10^4 <= target <= 10^4

這個題目跟 15 題非常相似,同樣是給定一組整數陣列輸入,不過還多給了一個目標值 target。我們需要找出 3 個數值,使其加總值越接近 target 越好。

需要注意的是,我們要返回的並不是找到的陣列,而是直接返回最接近 target 的數值。

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LeetCode: 15-3Sum 解題紀錄

Last Updated on 2021-06-02 by Clay


題目

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Notice that the solution set must not contain duplicate triplets.

Example:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Input: nums = []
Output: []

題目是給定一個全是由『整數』組成的陣列,而我們的目標就是要找出『任意三個數值加總為 0 的所有答案』,也是題目 3Sum 的意義。

還有,返回的結果不能有重複的答案。

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[Linux] 升級 sudo 套件的幾個方法

Last Updated on 2021-01-28 by Clay

Linux is an operating system that can do all the functions you imagine

前言

最近看到新聞,發現現在在 Unix-liked 系統中大家常用的 sudo 工具,原來存在著一個『堆積緩衝溢位漏洞』的 BUG(編號 CVE-2021-3156)。

目前在以下版本確認了能夠取得完整的 root 權限:

  • Ubuntu 20.04(sudo 1.8.31)
  • Debian 10(sudo 1.8.27)
  • Fedora 33(sudo 1.9.2)

而據報導、以及我粗略看過的幾個討論區,幾乎都在建議大家迅速升級該套件版本。這個 BUG 在 2021/01 月中發現而通報 sudo 開發單位,差不多一個禮拜多,新版的修復就已經上線了。

以下我簡單紀錄該如何升級 sudo 版本。

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LeetCode: 31-Next Permutation 解題紀錄

Last Updated on 2021-01-31 by Clay


題目

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

The replacement must be in place and use only constant extra memory.

Example:

Input: nums = [1,2,3]
Output: [1,3,2]

Input: nums = [3,2,1]
Output: [1,2,3]

Input: nums = [1,1,5]
Output: [1,5,1]

Input: nums = [1]
Output: [1]

題目給定輸入一個陣列,我們要將陣列排列成『下一個較大的值』。比方說題目輸入了 [1,2,3],那麼我們就不能排列成 [3,1,2]、而是要排列成 [1,3,2] —— 因為 132 才下一個較大的值,而非 312。

而當題目給定的陣列沒有下一個較大的排列時,則將陣列『從小排到大』。

順帶一提這題不會返回任何值,直接處理輸入陣列即可。

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LeetCode: 21-Merge Two Sorted Lists 解題紀錄

Last Updated on 2021-01-31 by Clay


題目

Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.

Example:

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

題目輸入兩個已經排序過的 Linked List,而我們要將其『合併』,並回傳同樣排序過的 Linked List。

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LeetCode: 13-Roman to Integer 解題紀錄

Last Updated on 2021-01-16 by Clay


題目

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000
For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
  • I can be placed before V (5) and X (10) to make 4 and 9. 
  • X can be placed before L (50) and C (100) to make 40 and 90. 
  • C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.

Example:

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
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