1、js 統計一個字符串出現頻率最高的字母/數字

let str = 'asdfghjklaqwertyuiopiaia'; const strChar = str => {let string = [...str],maxValue = '',obj = {},max = 0;string.forEach(value => {obj[value] = obj[value] == undefined ? 1 : obj[value] + 1if (obj[value] > max) {max = obj[value]maxValue = value}}) return maxValue; } console.log(strChar(str)) // a 復制代碼

2、數組去重

2.1、forEach

let arr = ['1', '2', '3', '1', 'a', 'b', 'b'] const unique = arr => {let obj = {}arr.forEach(value => {obj[value] = 0})return Object.keys(obj) } console.log(unique(arr)) // ['1','2','3','a','b'] 復制代碼

2.2、filter

let arr = ['1', '2', '3', '1', 'a', 'b', 'b'] const unique = arr => {return arr.filter((ele, index, array) => {return index === array.indexOf(ele)}) } console.log(unique(arr)) // ['1','2','3','a','b'] 復制代碼

2.3、set

let arr = ['1', '2', '3', '1', 'a', 'b', 'b'] const unique = arr => {return [...new Set(arr)] } console.log(unique(arr)) // ['1','2','3','a','b'] 復制代碼

2.4、reduce

let arr = ['1', '2', '3', '1', 'a', 'b', 'b'] const unique = arr.reduce((map, item) => {map[item] = 0return map }, {}) console.log(Object.keys(unique)) // ['1','2','3','a','b'] 復制代碼

3、翻轉字符串

let str ="Hello Dog"; const reverseString = str =>{return [...str].reverse().join(""); } console.log(reverseString(str)) // goD olleH 復制代碼

4、數組中最大差值

4.1、forEach

let arr = [23, 4, 5, 2, 4, 5, 6, 6, 71, -3]; const difference = arr => {let min = arr[0],max = 0;arr.forEach(value => {if (value < min) min = valueif (value > max) max = value})return max - min ; } console.log(difference(arr)) // 74 復制代碼

4.2、max、min

let arr = [23, 4, 5, 2, 4, 5, 6, 6, 71, -3]; const difference = arr => {let max = Math.max(...arr),min = Math.min(...arr);return max - min ; } console.log(difference(arr)) // 74 復制代碼

5、不借助臨時變量，進行兩個整數的交換

5.1、數組解構

let a = 2,b = 3;[b,a] = [a,b]console.log(a,b) // 3 2 復制代碼

5.2、算術運算（加減）

輸入a = 2,b = 3,輸出 a = 3,b = 2

let a = 2,b = 3; const swop = (a, b) => {b = b - a;a = a + b;b = a - b;return [a,b]; } console.log(swop(2,3)) // [3,2] 復制代碼

5.3、邏輯運算（異或）

let a = 2,b = 3; const swop = (a, b) => {a ^= b; //x先存x和y兩者的信息b ^= a; //保持x不變，利用x異或反轉y的原始值使其等于x的原始值a ^= b; //保持y不變，利用x異或反轉y的原始值使其等于y的原始值return [a,b]; } console.log(swop(2,3)) // [3,2] 復制代碼

6、排序 (從小到大)

6.1、冒泡排序

let arr = [43, 32, 1, 5, 9, 22]; const sort = arr => {let res = []arr.forEach((v, i) => {for (let j = i + 1; j < arr.length; j++) {if (arr[i] > arr[j]) {[arr[i],arr[j]] = [arr[j],arr[i]]}}})return arr } console.log(sort(arr)) // [1, 5, 9, 22, 32, 43] 復制代碼

結語

持續更新，持續更新...

錯漏之處，敬請斧正

有更多更好的js算法面試題,還望伙伴們積極推薦

轉載于:https://juejin.im/post/5a7aaf745188257a5a4c9a39

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