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Algorithms that are easy to encounter in JS interviews

May 26, 2018 pm 02:20 PM
javascript algorithm interview

Algorithms that are easy to encounter in JS interviews

This time I will bring you algorithms that are easy to encounter in JS interviews. The following is a practical case, let’s take a look.

Recommended related articles:The most complete collection of js interview questions in 2020 (latest)

Prime numbers

Q: How would you verify a prime number?

A: A prime number can only be divisible by itself and 1. So, I'm going to run a while loop and add 1. (Look at the code examples. If you can’t understand it, it’s not your cup of tea. Go back and learn the javaScript basics first and then come back.)

Method 1

function isPrime(n){
 var pisor = 2;
 while (n > pisor){
 if(n % pisor == 0){
  return false; 
 }
 else
  pisor++;
 }
 return true;
}
isPrime(137); // = true
isPrime(237); // = false
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Q: Can you do better?

A: Yes. The divisor increases by one at a time. After 3 I can add 2. If a number is divisible by any even number, it will be divisible by 2.

Supplement: If you don’t need to increase the divisor to this number. You can stop earlier. Let me explain it in the steps below (you can read it a few more times if you need to)

Step one, no number is divisible by a number greater than half of it. For example, 13 will never be divisible by 7,8,9...it can even be half of an even number. For example, 16 will be divisible by 8, but never by 9, 10, 11, 12...
Conclusion: A number will never be divisible by a number greater than half its value. So, we can loop 50% less.

Second step, now, if a number is not divisible by 3. (If it's divisible by 3, then it's not prime). Then, it is not divisible by any number greater than 1/3 of its value. For example, 35 is not divisible by 3. Therefore, it will never be divisible by a number greater than 35/3, never divisible by 12, 13, 14... If you take an even number like 36, it will never be divisible by 13, 14, 15.

Conclusion: A number is divisible by one-third of its value.

Step 3, for example, you have a number 127. 127 is not divisible by 2, so you should check at most 63.5. Secondly, 127 is not divisible by 3. So you will check to 127/3 approximately 42. It is not divisible by 5, the divisor should be less than 127/5 which is about 25, not 7. So, where do we stop?
Conclusion: The divisor will be less than math.sqrt(N)

Method 2

Don’t worry if you can’t understand it, just leave it alone. It doesn't matter if you're not a researcher.

function isPrime(n)
{
 var pisor = 3, 
  limit = Math.sqrt(n);
 //check simple cases
 if (n == 2 || n == 3)
  return true;
 if (n % 2 == 0)
  return false;
 while (pisor <= limit)
 {
 if (n % pisor == 0)
  return false;
 else
  pisor += 2;
 }
 return true;
}
isPrime(137); // = true
isPrime(237); // = false
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Prime factors

Q: How to find all the prime factors of a number?
A: Execute a while loop. Start dividing by 2. If it cannot be divided, record the divisor until completed.

function primeFactors(n){
 var factors = [], 
  pisor = 2;
 while(n>2){
 if(n % pisor == 0){
  factors.push(pisor); 
  n= n/ pisor;
 }
 else{
  pisor++;
 }  
 }
 return factors;
}
 primeFactors(69); // = [3, 23]
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Q: What is the running time complexity? Can you do better?

A: O(n). You can start the divisor from 3 and add up to 2. Because, if a number is divisible by any even number, it will be divisible by 2. Therefore, you don't need to divide by even numbers. Furthermore, you won't have a factor larger than half its value. If you want to make it complicated, use the supplementary concepts from the first question.

Fibonacci (Fibonacci)

Q: How to get the nth Fibonacci number?
A: I create an array and start by iterating.

The Fibonacci series is one of the most popular interview questions for beginners. So, you have to learn this one.

Method 1

function fibonacci(n){
 var fibo = [0, 1];
 if (n <= 2) return 1;
 for (var i = 2; i <=n; i++ ){
 fibo[i] = fibo[i-1]+fibo[i-2];
 }
 return fibo[n];
} 
fibonacci(12); // = 144
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Q: What is the running time complexity?

A: O(n);

Q: Can you make it recursive?

Method 2

function fibonacci(n){
 if(n < =1) {
  return n;
 } else {
  return fibonacci(n-1) + fibonacci (n-2);
 }
}
fibonacci(12); // = 144
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Q: What is the running time complexity?
A: O(2n); Details about time complexity

Greatest common divisor

Q: How would you find two What is the greatest common divisor of numbers?

function greatestCommonpisor(a, b){
 var pisor = 2, 
  greatestpisor = 1;
 //if u pass a -ve number this will not work. fix it dude!!
 if (a < 2 || b < 2)
  return 1;
 while(a >= pisor && b >= pisor){
 if(a %pisor == 0 && b% pisor ==0){
  greatestpisor = pisor;  
 }
 pisor++;
 }
 return greatestpisor;
}
greatestCommonpisor(14, 21); // 7
greatestCommonpisor(69, 169); // = 1
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Algorithm Paradigm

很抱歉。我也无法解释它。 因为我自己80%的情况下都不能理解它。 我的算法分析教练告诉我这个,又从课堂笔记偷走(我是一个好学生,顺便说一句!)

function greatestCommonpisor(a, b){
 if(b == 0)
  return a;
 else 
  return greatestCommonpisor(b, a%b);
}
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注意:用你的大脑来理解它。

去重

Q:你将如何从数组中删除重复的成员?
A: 执行一个循环,并保存一个对象/关联数组。如果我第一次找到一个元素,我会设置它的值为真(这将告诉我元素添加一次)。如果我在对象中找到这个元素,我不会将它插入到返回数组中。

function removeDuplicate(arr){
 var exists ={},
  outArr = [], 
  elm;
 for(var i =0; i<arr.length; i++){
 elm = arr[i];
 if(!exists[elm]){
  outArr.push(elm);
  exists[elm] = true;
 }
 }
 return outArr;
}
removeDuplicate([1,3,3,3,1,5,6,7,8,1]); // = [1, 3, 5, 6, 7, 8]
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合并两个排序的数组

Q: 怎样合并两个已排序数组?
A: 我将为每个数组保留一个指针(看代码,并注意这个)。

function mergeSortedArray(a, b){
 var merged = [], 
   aElm = a[0],
   bElm = b[0],
   i = 1,
   j = 1;
 if(a.length ==0)
  return b;
 if(b.length ==0)
  return a;
 /* 
 if aElm or bElm exists we will insert to merged array
 (will go inside while loop)
  to insert: aElm exists and bElm doesn&#39;t exists
       or both exists and aElm < bElm
  this is the critical part of the example      
 */
 while(aElm || bElm){
  if((aElm && !bElm) || aElm < bElm){
   merged.push(aElm);
   aElm = a[i++];
  }  
  else {
   merged.push(bElm);
   bElm = b[j++];
  }
 }
 return merged;
}
mergeSortedArray([2,5,6,9], [1,2,3,29]);// = [1, 2, 2, 3, 5, 6, 9, 29]
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不通过临时变量交换两个数的值

Q:如何在不使用临时变量的情况下交换两个数字?

function swapNumb(a, b){
 console.log(&#39;before swap: &#39;,&#39;a: &#39;, a, &#39;b: &#39;, b);
 b = b -a;
 a = a+ b;
 b = a-b;
 console.log(&#39;after swap: &#39;,&#39;a: &#39;, a, &#39;b: &#39;, b); 
}
swapNumb(2, 3);
//  = before swap: a: 2 b: 3
//  = after swap: a: 3 b: 2
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位操作:对不起,我无法向你解释这一点。 Kinjal Dave建议到 logical conjunction理解它。将浪费您30分钟。

function swapNumb(a, b){
 console.log("a: " + a + " and b: " + b);
 a = a ^ b;
 b = a ^ b;
 a = a ^ b;
 console.log("a: " + a + " and b: " + b);
}
swapNumb(2, 3);
// = a: 2 and b: 3
// = a: 3 and b: 2
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字符串反向

Q:如何在JavaScript中反转字符串?
A:可以遍历字符串并将字母连接到新字符串。

方法1

function reverse(str){
 var rtnStr = &#39;&#39;;
 for(var i = str.length-1; i>=0;i--){
  rtnStr +=str[i];
 }
 return rtnStr;
}
reverse('you are a nice dude');
// = "edud ecin a era uoy"
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Q:你知道在现代浏览器中串联效果很好,但在像IE8这样的旧浏览器中会变慢。 还有什么不同的方法,可以扭转一个字符串?

A:当然.我可以使用数组,也可以添加一些检查。如果字符串是NULL或其他字符串,这将失败。让我也做一些类型检查。使用此数组类似于在某些服务器端语言中使用字符串缓冲区。

方法2

function reverse(str){
 var rtnStr = [];
 if(!str || typeof str != 'string' || str.length < 2 ) return str;
 for(var i = str.length-1; i>=0;i--){
  rtnStr.push(str[i]);
 }
 return rtnStr.join('');
}
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Q: 运行时间复杂度是多少?
A: O(n);
Q:可以做得更好?
A:我可以遍历索引的一半,它会节省一点点。 (这是没用的,可能不会打动面试官)

方法3

function reverse(str) {
 str = str.split('');
 var len = str.length,
   halfIndex = Math.floor(len / 2) - 1,
   revStr;
 for (var i = 0; i <= halfIndex; i++) {
  revStr = str[len - i - 1];
  str[len - i - 1] = str[i];
  str[i] = revStr;
 }
 return str.join(&#39;&#39;);
}
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Q:这有效,但你可以做递归的方式吗?
A:可以。

方法4

function reverse (str) {
  if (str === "") {
    return "";
  } else {
    return reverse(str.substr(1)) + str.charAt(0);
  }
}
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方法5

Q:你可以在方法中使用任何构建,使它更清洁吗?

function reverse(str){
 if(!str || str.length <2) return str;
 return str.split(&#39;&#39;).reverse().join(&#39;&#39;);
}
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方法6

Q:你可以做反向函数作为字符串扩展吗?
A:我需要将这个函数添加到String.prototype,而不是使用str作为参数,我需要使用this

String.prototype.reverse = function (){
 if(!this || this.length <2) return this;
 return this.split(&#39;&#39;).reverse().join(&#39;&#39;);
}
&#39;abc&#39;.reverse();
// = &#39;cba&#39;
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单词反转

Q:你如何在句子中颠倒单词?
A:您必须检查整个字符串的空白区域。确定是否可能有多个空格。

//have a tailing white space
//fix this later
//now i m sleepy
function reverseWords(str){
 var rev = [], 
   wordLen = 0;
 for(var i = str.length-1; i>=0; i--){
  if(str[i]==' ' || i==0){
   rev.push(str.substr(i,wordLen+1));
   wordLen = 0;
  }
  else
   wordLen++;
 }
 return rev.join(' ');
}
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内置方法的快速解决方案:

function reverseWords(str){
 return str.split(' ').reverse();
}
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原位反转

Q: 如果你有一个字符串如”I am the good boy”, 怎样变为 “I ma eht doog yob”? 注意这些单词位置不变但是被反转了。

A: 要做到这一点,我必须做字符串反向和字反转。

function reverseInPlace(str){
 return str.split(' ').reverse().join(' ').split('').reverse().join('');
}
reverseInPlace('I am the good boy');// = "I ma eht doog yob"
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Q: ok。好的,你能不使用内置反向函数做到吗?
A: (内心独白)有没有搞错!!

//sum two methods.
//you can simply split words by ' '
//and for each words, call reverse function
//put reverse in a separate function
//if u cant do this, 
//have a glass of water, and sleep
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第一个非重复字符

Q: 怎么在字符串中找到第一个非重复字符?
A: 有什么条件吗?
A: 比如是否区分大小写?
面试官可能会说No。
A: 是长字符串还是短字符串?
Q: 这些有什么关系吗?

A:例如,如果它是一个非常长的字符串,说一百万个字符,我想检查是否有26个英文字符正在重复。 我可能会检查是否所有字符都在每200个字母中重复(例如),而不是循环遍历整个字符串。 这将节省计算时间。
Q: 简单起见, 这个字符串是 “the quick brown fox jumps then quickly blow air”。

function firstNonRepeatChar(str){
 var len = str.length,
   char, 
   charCount = {};
 for(var i =0; i<len; i++){
  char = str[i];
  if(charCount[char]){
   charCount[char]++;
  }
  else
   charCount[char] = 1;
 }
 for (var j in charCount){
  if (charCount[j]==1)
    return j;
 }
} 
firstNonRepeatChar(&#39;the quick brown fox jumps then quickly blow air&#39;);// = "f"
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这有一个问题,不能再循环中及时退出。

删除重复的字符

Q: 怎样删除字符串中的重复字符?

A: 这与第一个非重复字符非常相似。你应该问一些类似的问题。它是区分大小写的吗?。

如果面试官说,这是区分大小写的,那么你就很轻松了。 如果他说不。你可以使用string.toLowercase()来把字符串。面试官可能不喜欢这个方法。 因为返回字符串不会拥有相同的大小写。 所以

function removeDuplicateChar(str){
 var len = str.length,
   char, 
   charCount = {}, 
   newStr = [];
 for(var i =0; i<len; i++){
  char = str[i];
  if(charCount[char]){
   charCount[char]++;
  }
  else
   charCount[char] = 1;
 }
 for (var j in charCount){
  if (charCount[j]==1)
    newStr.push(j);
 }
 return newStr.join(&#39;&#39;);
}
removeDuplicateChar(&#39;Learn more javascript dude&#39;); // = "Lnmojvsciptu"
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回文检查

Q: 如何检查一个字符串是否是回文?

A: 把字符串反转,如果反转前后相等,那么它就是回文。

function isPalindrome(str){
 var i, len = str.length;
 for(i =0; i<len/2; i++){
  if (str[i]!== str[len -1 -i])
    return false;
 }
 return true;
}
isPalindrome(&#39;madam&#39;)
// = true
isPalindrome(&#39;toyota&#39;)
// = false
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或者

function checkPalindrom(str) {
  return str == str.split(&#39;&#39;).reverse().join(&#39;&#39;);
}
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类似的:在 O(n)时间复杂度内判断一个字符串是否包含在回文字符串内。你能在O(1)时间解决问题吗?

找缺失的数字

Q: 在一个1到100的未排序数组中找到缺失的数,你怎么做?

说明:数组中的数字为1到100。 数组中只有一个数字缺失。数组未排序。找到缺少的数字。

A: 你必须表现得像是在想很多。然后讨论n=n(n+1)/2的线性级数之和

function missingNumber(arr){
 var n = arr.length+1, 
   sum = 0,
   expectedSum = n * (n+1)/2;
 for(var i = 0, len = arr.length; i < len; i++){
  sum += arr[i];
 }
 return expectedSum - sum;
}
missingNumber([5, 2, 6, 1, 3]);
// = 4
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注意: 这个会返回任意长度数组中缺失的那个

两数之和

Q: 在一个未排序的数组中找出是否有任意两数之和等于给定的数?
A: 简单!双重循环。

function sumFinder(arr, sum){
 var len = arr.length;
 for(var i =0; i<len-1; i++){ 
   for(var j = i+1;j<len; j++){
    if (arr[i] + arr[j] == sum)
      return true;
   }
 }
 return false;
}
sumFinder([6,4,3,2,1,7], 9);
// = true
sumFinder([6,4,3,2,1,7], 2);
// = false
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Q: 时间复杂度?
A: O(n2)。
Q: 有更优解?
A: 我想想。我可以用一个对象来存储当前元素和和值的差值。当我拿到一个新元素,如果这个元素的差值在对象中存在,那么我就能判断出是否存在。

function sumFinder(arr, sum){
 var differ = {}, 
   len = arr.length,
   substract;
 for(var i =0; i<len; i++){
   substract = sum - arr[i];
   if(differ[substract])
    return true;    
   else
    differ[arr[i]] = true;
 }
 return false;
}
sumFinder([6,4,3,2,1,7], 9);
// = true
sumFinder([6,4,3,2,1,7], 2);
// = false
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最大和

Q: 找到任意两个元素的最大总和?

A: 这实际上非常简单直接。 找到两个最大的数字并返回它们的总和

function topSum(arr){
 var biggest = arr[0], 
   second = arr[1], 
   len = arr.length, 
   i = 2;
 if (len<2) return null;
 if (biggest<second){
  biggest = arr[1];
  second = arr[0];
 } 
 for(; i<len; i++){
  if(arr[i] > biggest){
   second = biggest;
   biggest = arr[i];
  }
  else if (arr[i]>second){
   second = arr[i];
  }
 }
 return biggest + second;
}
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统计零

Q: 统计从1到n的零总数?

A: 如果 n = 100,则0的数目将是11(0,10,20,30,40,50,60,70,80,90,100)。 请注意,100有两个0.这个看起来很简单,但有点棘手

说明:所以这里的重点是。 如果你有一个1到50的数字,那么这个数值就是5,就是50除以10.然而,如果这个数值是100,这个数值是11,你将得到100/10 = 10和 10/10 = 1。 那就是你将如何在一个数字中得到更多的零,如(100,200,1000);

function countZero(n){
 var count = 0;
 while(n>0){
  count += Math.floor(n/10);
  n = n/10;
 }
 return count;
}
countZero(2014);
// = 223
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子字符串

Q: 在字符串中匹配子字符串?

A: 在迭代字符串时将使用指针(一个用于字符串,另一个用于子字符串)。 然后用另一个变量来保存初始匹配的起始索引。

function subStringFinder(str, subStr){
 var idx = 0,
   i = 0,
   j = 0,
   len = str.length,
   subLen = subStr.length;
  for(; i<len; i++){
   if(str[i] == subStr[j])
     j++;
   else
     j = 0;
   //check starting point or a match  
   if(j == 0)
    idx = i;
   else if (j == subLen)
    return idx;
 }
 return -1;
}
subStringFinder('abbcdabbbbbck', 'ab')
// = 0
subStringFinder('abbcdabbbbbck', 'bck')
// = 9
//doesn't work for this one.
subStringFinder('abbcdabbbbbck', 'bbbck') 
// = -1
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排列

Q: 如何获取字符串中的所有排列?

A: 根据您对算法的了解程度,这可能会很困难。、

function permutations(str){
  var arr = str.split(''),
    len = arr.length, 
    perms = [],
    rest,
    picked,
    restPerms,
    next;
  if (len == 0)
    return [str];
  for (var i=0; i<len; i++)
  {
    rest = Object.create(arr);
    picked = rest.splice(i, 1);
    restPerms = permutations(rest.join(''));
    for (var j=0, jLen = restPerms.length; j< jLen; j++)
    {
      next = picked.concat(restPerms[j]);
      perms.push(next.join(''));
    }
  }
  return perms;
}
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SublimeText3 Mac version

God-level code editing software (SublimeText3)

CLIP-BEVFormer: Explicitly supervise the BEVFormer structure to improve long-tail detection performance CLIP-BEVFormer: Explicitly supervise the BEVFormer structure to improve long-tail detection performance Mar 26, 2024 pm 12:41 PM

Written above &amp; the author’s personal understanding: At present, in the entire autonomous driving system, the perception module plays a vital role. The autonomous vehicle driving on the road can only obtain accurate perception results through the perception module. The downstream regulation and control module in the autonomous driving system makes timely and correct judgments and behavioral decisions. Currently, cars with autonomous driving functions are usually equipped with a variety of data information sensors including surround-view camera sensors, lidar sensors, and millimeter-wave radar sensors to collect information in different modalities to achieve accurate perception tasks. The BEV perception algorithm based on pure vision is favored by the industry because of its low hardware cost and easy deployment, and its output results can be easily applied to various downstream tasks.

Explore the underlying principles and algorithm selection of the C++sort function Explore the underlying principles and algorithm selection of the C++sort function Apr 02, 2024 pm 05:36 PM

The bottom layer of the C++sort function uses merge sort, its complexity is O(nlogn), and provides different sorting algorithm choices, including quick sort, heap sort and stable sort.

Implementing Machine Learning Algorithms in C++: Common Challenges and Solutions Implementing Machine Learning Algorithms in C++: Common Challenges and Solutions Jun 03, 2024 pm 01:25 PM

Common challenges faced by machine learning algorithms in C++ include memory management, multi-threading, performance optimization, and maintainability. Solutions include using smart pointers, modern threading libraries, SIMD instructions and third-party libraries, as well as following coding style guidelines and using automation tools. Practical cases show how to use the Eigen library to implement linear regression algorithms, effectively manage memory and use high-performance matrix operations.

Can artificial intelligence predict crime? Explore CrimeGPT's capabilities Can artificial intelligence predict crime? Explore CrimeGPT's capabilities Mar 22, 2024 pm 10:10 PM

The convergence of artificial intelligence (AI) and law enforcement opens up new possibilities for crime prevention and detection. The predictive capabilities of artificial intelligence are widely used in systems such as CrimeGPT (Crime Prediction Technology) to predict criminal activities. This article explores the potential of artificial intelligence in crime prediction, its current applications, the challenges it faces, and the possible ethical implications of the technology. Artificial Intelligence and Crime Prediction: The Basics CrimeGPT uses machine learning algorithms to analyze large data sets, identifying patterns that can predict where and when crimes are likely to occur. These data sets include historical crime statistics, demographic information, economic indicators, weather patterns, and more. By identifying trends that human analysts might miss, artificial intelligence can empower law enforcement agencies

Improved detection algorithm: for target detection in high-resolution optical remote sensing images Improved detection algorithm: for target detection in high-resolution optical remote sensing images Jun 06, 2024 pm 12:33 PM

01 Outlook Summary Currently, it is difficult to achieve an appropriate balance between detection efficiency and detection results. We have developed an enhanced YOLOv5 algorithm for target detection in high-resolution optical remote sensing images, using multi-layer feature pyramids, multi-detection head strategies and hybrid attention modules to improve the effect of the target detection network in optical remote sensing images. According to the SIMD data set, the mAP of the new algorithm is 2.2% better than YOLOv5 and 8.48% better than YOLOX, achieving a better balance between detection results and speed. 02 Background & Motivation With the rapid development of remote sensing technology, high-resolution optical remote sensing images have been used to describe many objects on the earth’s surface, including aircraft, cars, buildings, etc. Object detection in the interpretation of remote sensing images

Application of algorithms in the construction of 58 portrait platform Application of algorithms in the construction of 58 portrait platform May 09, 2024 am 09:01 AM

1. Background of the Construction of 58 Portraits Platform First of all, I would like to share with you the background of the construction of the 58 Portrait Platform. 1. The traditional thinking of the traditional profiling platform is no longer enough. Building a user profiling platform relies on data warehouse modeling capabilities to integrate data from multiple business lines to build accurate user portraits; it also requires data mining to understand user behavior, interests and needs, and provide algorithms. side capabilities; finally, it also needs to have data platform capabilities to efficiently store, query and share user profile data and provide profile services. The main difference between a self-built business profiling platform and a middle-office profiling platform is that the self-built profiling platform serves a single business line and can be customized on demand; the mid-office platform serves multiple business lines, has complex modeling, and provides more general capabilities. 2.58 User portraits of the background of Zhongtai portrait construction

Add SOTA in real time and skyrocket! FastOcc: Faster inference and deployment-friendly Occ algorithm is here! Add SOTA in real time and skyrocket! FastOcc: Faster inference and deployment-friendly Occ algorithm is here! Mar 14, 2024 pm 11:50 PM

Written above & The author’s personal understanding is that in the autonomous driving system, the perception task is a crucial component of the entire autonomous driving system. The main goal of the perception task is to enable autonomous vehicles to understand and perceive surrounding environmental elements, such as vehicles driving on the road, pedestrians on the roadside, obstacles encountered during driving, traffic signs on the road, etc., thereby helping downstream modules Make correct and reasonable decisions and actions. A vehicle with self-driving capabilities is usually equipped with different types of information collection sensors, such as surround-view camera sensors, lidar sensors, millimeter-wave radar sensors, etc., to ensure that the self-driving vehicle can accurately perceive and understand surrounding environment elements. , enabling autonomous vehicles to make correct decisions during autonomous driving. Head

Golang framework interview questions collection Golang framework interview questions collection Jun 02, 2024 pm 09:37 PM

The Go framework is a set of components that extend Go's built-in libraries, providing pre-built functionality (such as web development and database operations). Popular Go frameworks include Gin (web development), GORM (database operations), and RESTful (API management). Middleware is an interceptor pattern in the HTTP request processing chain and is used to add functionality such as authentication or request logging without modifying the handler. Session management maintains session status by storing user data. You can use gorilla/sessions to manage sessions.

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