30Day LeetCoding Callenge — Week1
30-Day Leecodeing Callenge ,第一週挑戰 !!
藉由此活動,我想更深入去瞭解JavaScript的演算法思維,目標是訓練出腦海中能觀想出演算法模型。
總結這週的技巧,有兩個值得注意的演算法。
1.Hashmap
javascript 的 hashmap 就是應用Object的索引值存跨索引值得值
可參考
大致觀念是這樣:
let arr = ['a','b','c'] let hashmap = {} for(let i = 0; i < arr.length ; i++){ if(arr[i] = 'a') arr[arr[i]] = 1 } // hashmap = { {'a': 1} }
2.two pointers
就是從陣列兩端開始走訪,直到兩端交錯
可參考:
GreeksforGreeks — Two Pointers Technique
let i = 0 let j = arr.length while(i < j){ if(condition){ dosomething } i++ j-- }
題目
1.Single Number
說明:找出陣列中不重複的數字
Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
思維:
我的解法是先累計每個數字的數目,然後再找數目唯一的數字
先用Object 紀錄最後再轉Array 處理,算是蠻常用到的技巧
Answer:
var singleNumber = function(nums) { let temp = {} nums.map(num => { if(!temp[num]){ temp[num] = 1 }else{ temp[num] +=1 } }) return Object.keys(temp).find(key => 1 === temp[key]) };
2. Happy Number
說明:判斷是否為快樂數
Example:
Input: 19 Output: true Explanation: 12 + 92 = 82 82 + 22 = 68 62 + 82 = 100 12 + 02 + 02 = 1
思維:
- 歸納出不快樂數比對輸入數字
- 判斷數字拆開相乘的數是不是快樂數,然後回到1.判斷是否快樂
Answer:
var isHappy = function(n) { let answer = false let notHappy = [4,,16,37,58,89,145,42,20] let num = n while(notHappy.indexOf(num) === -1){ if(num ===1){ answer = true break }else{ let list = num.toString().split("") let sum = 0 let newNum = list.map((a)=>{ return sum += Math.pow(a,2) }) num = sum } } return answer };
3.Maximum Subarray
說明:找出陣列中連續加總最大和
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
思維:
這題目有點抽象,首先先找出最大值,然後加總為負數則設為0
我後來想想如果最大和是負數該怎麼辦?如果是這樣的話那最大值應該就是最大和了,
所以小於0就等於0比較像是連續陣列的斷點,用這樣想比較好懂
Answer:
var maxSubArray = function(nums) { let max = Math.max(...nums) let num = 0 for(let i = 0 ; i < nums.length ; i++){ num += nums[i] if(num > max) max = num if(num < 0) num = 0 } return max };
4.Move Zeroes
說明:把所有0移到最後面
Example:
Input: [0,1,0,3,12] Output: [1,3,12,0,0]
思維:
題目有提示適合用two pointer , two pointer 是簡單好用的演算法
看題目就可以知道,這個演算法就是從兩頭開始比對直到交錯為止,
陣列右邊(尾端)是放0得地方,所以遇到0就向左走
陣列左邊(頭端)是移動0的地方,所以用splice(i,1)切掉0,在push(0)
Answer:
var moveZeroes = function(nums) { let i = 0 let j = nums.length - 1 while(i < j ){ if(nums[j] === 0) j-- if(nums[i] === 0) { nums.splice(i,1) nums.push(0) }else{ i++ } } };
5.Best Time to Buy and Sell Stock II
說明:股票陣列中找出最大收益和
Example 1:
Input: [7,1,5,3,6,4] Output: 7 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Example 2:
Input: [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
思維:
這題沒有很複雜,題目有說明都是單次交易所以只要當天比隔天還低就是賺
Answer:
var maxProfit = function(prices) { let i = 0 let j = prices.length let sumProfit = 0 while(i < j-1 ){ let profit = 0 if(prices[i] < prices[i+1] ){ profit = prices[i+1]- prices[i] } sumProfit += profit i++ } return sumProfit };
6.Group Anagrams
說明:找出字母一樣的單字並分類
Example:
Input: ["eat", "tea", "tan", "ate", "nat", "bat"], Output: [ ["ate","eat","tea"], ["nat","tan"], ["bat"] ]
思維:
這題難在於輸入跟輸出的陣列是非對稱的,所以有時要嘗試 hashmap
string.split(‘’).sort()則是用來判斷是否同類
Answer:
var groupAnagrams = function(strs) { let hash = { } strs.map(s =>{ let temp = "" let math = s.split('').sort() if(temp !== math) temp = math if(temp === math){ if(hash[math]){ hash[math].push(s) }else{ hash[math] = [s] } } }) return (Object.values(hash)) };
7.Counting Elements
說明:計算陣列中存在 n+1 的數。
Example 1:
Input: arr = [1,2,3] Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
Example 2:
Input: arr = [1,1,3,3,5,5,7,7] Output: 0 Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.
Example 3:
Input: arr = [1,3,2,3,5,0] Output: 3 Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
Example 4:
Input: arr = [1,1,2,2] Output: 2 Explanation: Two 1s are counted cause 2 is in arr.
思維:hashmap的變形應用
- 找出有沒有n+1的數,並累計有幾個
- 再找一次陣列中的值有幾個符合n+1
- 累加hashmap裡面符合n+1的數量
Answer:
var countElements = function(arr) { let hash = {} arr.map(num => { if(!hash[num +1 ]){ hash[num +1 ] = 1 } else{ hash[num +1 ] += 1 } }) let index = Object.keys(hash).filter((h,i) =>{ return arr.some(a =>{ return a == h }) }) let answer = 0 index.map((i) =>{ answer += hash[i] }) return answer };
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