一、

Description:

Write a small function that returns the values of an array that are not odd.

All values in the array will be integers. Return the good values in the order they are given.

my solution:

function noOdds(values) {    var arr = [];    values.filter(function (val) {        if (val % 2 === 0) {            arr.push(val);        }    });    console.log(arr);    return arr;}

solution from web:

function no_odds( values ){  // Return all non-odd values  return values.filter(function(val){return val%2===0})}

二、

Description:

example:

getMissingElement( [0, 5, 1, 3, 2, 9, 7, 6, 4] ) // returns 8 getMissingElement( [9, 2, 4, 5, 7, 0, 8, 6, 1] ) // returns 3

mysolution:

//my solutionfunction getMissingElement(superImportantArray) {    var arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];    var result;    var i = 0;    for (i = 0; i < arr.length; i++) {        var exit = false;        for (var j = 0; j < superImportantArray.length; j++) {            if (arr[i] === superImportantArray[j]) {                exit = true;                break;            }        }        if (exit === false) {            result = arr[i];            break;        }    }    console.log(result);    return result;}

solution from web:

//外国佬的解决方案function getMissingElement(superImportantArray) {    return 45 - superImportantArray.reduce(function (a, b) {        return a + b;    });}

三、

Description:

Complete the keysAndValues function so that it takes in an object and returns the keys and values as separate arrays.

Example:

keysAndValues({a: 1, b: 2, c: 3}) // should return [['a', 'b', 'c'], [1, 2, 3]]

mysolution:

function keysAndValues(data) {    var keys = [];    var maps = [];    for (var index in data) {        keys.push(index);        maps.push(data[index]);    }    var datas = [keys, maps];    console.log(datas);    return datas;}keysAndValues({a: 1, b: 2, c: 3});

solution from web:

function keysAndValues(data){  var keys = Object.getOwnPropertyNames(data),      vals = keys.map(function (key) { return data[key]; });  return [keys, vals];}

四、

Description:

Happy Holidays fellow Code Warriors!

It's almost Christmas! That means Santa's making his list, and checking it twice. Unfortunately, Santa's Javascript and CoffeeScript Elves accidentally mixed the Naughty and Nice list together! Santa needs your help to save Christmas!

Save Christmas!

Santa needs you to write two functions, getNiceNames and getNaughtyNames. Both of the functions accept an array of objects. getNiceNames returns an array containing only the names of the nice people, and getNaughtyNames returns an array containing only the names of the naughty people. Return an empty array [] if the result from either of the functions contains no names.

The objects in the passed in array will represent people. Each object contains two properties: name and wasNice.

name - The name of the personwasNice - True if the person was nice this year, false if they were naughty

Person Object Examples:

{ name: 'Warrior reading this kata', wasNice: true } { name: 'xDranik', wasNice: false }

Test Example：

getNiceNames( [    { name: 'Warrior reading this kata', wasNice: true },    { name: 'xDranik', wasNice: false },    { name: 'Santa', wasNice: true }] )// => returns [ 'Warrior reading this kata', 'Santa' ]getNaughtyNames( [    { name: 'Warrior reading this kata', wasNice: true },    { name: 'xDranik', wasNice: false },    { name: 'Santa', wasNice: true }] )// => returns [ 'xDranik' ]

my solution：

function getNiceNames(people) {    var names = [];    for (var index in people) {        if (people[index].wasNice === true) {            names.push(people[index].name);        }    }    console.log(names);    return names;}function getNaughtyNames(people) {    var names = [];    for (var index in people) {        if (people[index].wasNice === false) {            names.push(people[index].name);        }    }    console.log(names);    return names;}

solution from web：

function getNiceNames(people){  return people.filter(function(e){    return e.wasNice;  }).map(function(e){    return e.name;  });}function getNaughtyNames(people){  return people.filter(function(e){    return !e.wasNice;  }).map(function(e){    return e.name;  });}

五、

Description:

my solution：

var isAnagram = function (test, original) {    var tests = test.toLowerCase().split("").sort();    var orgs = original.toLowerCase().split("").sort();    console.log(tests,orgs);    return tests.join(",") === orgs.join(",");};