Bubble sort, array de-weight, determine the number of occurrences of each value in the array ...

Don't worry about the road without bosom friend, the world who does not know June---Don Gao Shi "don't Dongda two first"

In this article, array sorting, array de-weight, judging the number of occurrences, do not use the method of the array.

The specific process thought in the code comment

First, bubble sort

//Bubble Sort:vararr = [1,2,3,5,4]; varJ,temp,flag; varCount =0;//number of wheels used to record comparisons//the number of times the outer loop control compares//Use assumption Method: (How to tag with variables)//1 Find an indeterminate operation in the program, set the assumptions//2 validation of assumptions (setting a condition that allows assumptions to fail)//3 Test The final value of the hypothetical condition, set the operation for(vari =0; I < arr.length-1; i++) {Count++; Flag=true;//Let's say you can sort of jump out of this round for(j =0; J < Arr.length-1I J + +){        if(Arr[j] > arr[j+1]) {temp=Arr[j]; ARR[J]= arr[j+1]; Arr[j+1] =temp; //as soon as you enter the IF, the ordering may not be complete, set the assumed condition to false//Stop JumpingFlag =false; }      }        if(flag) { Break; }} console.log (Arr,count);

First, the array to weight

//array de-weight:vararr = [5,1,2,1,2,2,1,2,1,2,3,4,3,4,5,3,4,5,4,3,2,1]; //the desired result is [1,2,3,4,5]//Console.log (Resultarr);//[1,2,3,4,5]//1 Creating an array of results varResultarr = [arr[0]]; varFlag; //2 times the enumeration group for(vari =1; i < arr.length; i++){      //the right way to achieveFlag =true;//assuming that the current arr[i] can be put into Resultarr for(j =0; J < Resultarr.length; J + +){        //set conditions that allow assumptions to fail: if you encounter any one of the elements equal, that means you cannot putif(Arr[i] = = =Resultarr[j]) {Flag=false; }      }      //set the final action based on the value of flagif(flag) {Resultarr.push (arr[i]); }} console.log (Resultarr);

简单介绍一下Set：Setis a new object in the ES6, which is especially convenient for the array. Only two lines of code are required

 //first defines a repeating array 
 var  arr = [1,1 , " Span style= "color: #800080;" >2 , 2,2 , 3,3 , 4,4 ] var  set  = new  Set (arr) //  {1,2,3,4}  var  newArr = Array. from  (set ) //  console.log (NEWARR)  //  [1,2,3,4]  

The principle of implementation: the difference is that Set Array Array duplicate elements are allowed in, for example, [1,2,2,3] and Set all elements in are unique and can only be {1,2,3} . With this feature, we can quickly remove duplicate elements from the array.

Iii. Number of occurrences

The first of these methods

The first type:

/*var arr = [ 5,1,2,1,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,2,2,1,2,1,2,3,4    , 5,3,4,5,4,3,2,1];    Want to know how many times the array of arr 1 appears?    var count = 0;      for (var i = 0; i < arr.length; i++) {if (arr[i] = = 1) {count++; }    }*/    //want to detect five of the data, the number of results obtained is also fivevarCountarr =[],count; //take out each data to be counted Arr[i] for(i =0; i < resultarr.length; i++) {Count=0;//Initialize the Count of Arr[i]//use Resultarr[i] to compare counts with each element in the original array arr  for(j =0; j<arr.length;j++){        if(Resultarr[i] = = =Arr[j]) {Count++; }      }      //Save the value of Count to CountarrCountarr.push (count); } console.log (Countarr);

The second method of

Bubble sort, array de-weight, determine the number of occurrences of each value in the array ...