Detailed explanation of Js interview algorithm-JS Tutorial-php.cn

Detailed explanation of Js interview algorithm

The rise of AI has made everyone pay more and more attention to algorithms. As a front-end engineer, algorithms are often our weakness. This article is translated from a foreign interview question. Listed some simple algorithm-related interview questions

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 thing. Go back and learn the basics of javaScript 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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Algorithmic Paradigm

Sorry. I can't explain it either. Because I myself can't understand it 80% of the time. My algorithm analysis instructor told me this, and stole it from class notes (I was a good student, btw!)

function greatestCommonpisor(a, b){
 if(b == 0)
  return a;
 else 
  return greatestCommonpisor(b, a%b);
}

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NOTE: Use your brain to understand it.

Deduplication

Q: How would you remove duplicate members from an array?
A: Execute a loop and save an object/associative array. If I find an element for the first time, I set its value to true (which will tell me the element was added once). If I find this element in the object, I don't insert it into the return array.

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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Merge two sorted arrays

Q: How to merge two sorted arrays?
A: I will keep a pointer for each array (look at the code and note this).

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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Exchanging the values of two numbers without using temporary variables

Q: How to exchange two numbers without using temporary variables?

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(&#39;you are a nice dude&#39;);
// = "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 != &#39;string&#39; || str.length < 2 ) return str;
 for(var i = str.length-1; i>=0;i--){
  rtnStr.push(str[i]);
 }
 return rtnStr.join(&#39;&#39;);
}

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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(&#39;&#39;);
 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]==&#39; &#39; || i==0){
   rev.push(str.substr(i,wordLen+1));
   wordLen = 0;
  }
  else
   wordLen++;
 }
 return rev.join(&#39; &#39;);
}

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内置方法的快速解决方案：

function reverseWords(str){
 return str.split(&#39; &#39;).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(&#39; &#39;).reverse().join(&#39; &#39;).split(&#39;&#39;).reverse().join(&#39;&#39;);
}
reverseInPlace(&#39;I am the good boy&#39;);// = "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 &#39; &#39;
//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(&#39;abbcdabbbbbck&#39;, &#39;ab&#39;)
// = 0
subStringFinder(&#39;abbcdabbbbbck&#39;, &#39;bck&#39;)
// = 9
//doesn&#39;t work for this one.
subStringFinder(&#39;abbcdabbbbbck&#39;, &#39;bbbck&#39;) 
// = -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(&#39;&#39;),
    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(&#39;&#39;));
    for (var j=0, jLen = restPerms.length; j< jLen; j++)
    {
      next = picked.concat(restPerms[j]);
      perms.push(next.join(&#39;&#39;));
    }
  }
  return perms;
}

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上面是我整理给大家的，希望今后会对大家有帮助。

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