bubble sort java

voidMain (string[] args) {5 int[] n = {0,5,2,6,9,3,4,8,1};6 inti,j;7 for(intx=1;x){8System.out.print (N[x] + ",");9 }TenSystem.out.println (""); One A for(i=1;i){ - for(j=n.length-1;j>=2;j--){ - if(n[j]]){ the inttemp =N[j]; -N[J] = n[j-1]; -N[J-1] =temp; - } + } - } + A for(intx=1;x) atSystem.out.print (n[x]+ ","); - } -}Three, the optimi

First, the basic idea:Bubble sort is a simple sort of exchange class. The basic idea is to scan the elements to be sorted from the beginning, compare the adjacent elements in sequence during the scanning process, and move the elements with large key values back. After each pass, the element with the highest key value is moved to the end, where the element's position is noted, and the next order only needs t

Principle: Assigning large values of adjacent elements to the rightIdea: ①1. Compares the first element in the collection or array to the second element and assigns a larger value to the right;2. Compare the second element to the third one, and the larger value to the right;.......(N-1). Compares the N-1 element with the nth element and assigns a larger value to the right;From this, the maximum value of the element in the set is obtained and assigned to the element n;② Repeat Operation ① get the

Simple Bubble Sort:ImportJava.lang.reflect.Array;Importjava.util.ArrayList;Importjava.util.Arrays; Public classTest {Static inta[]={5,2,11,4,31,1,6}; Public Static voidMain (string[]age) {System.out.println (arrays.tostring (a)); for(inti = 0; i ) { for(intj = 0; J ) { if(a[j]>a[j+1]) { intb=A[j]; A[J]=a[j+1]; A[j+1]=C; }}} System.out.println (Arrays.tostring (a)); }}

Bubble Sort Algorithm Demo diagram: public static void Bubblesort (int[] array) { //The next two options can be ///mode 1 for (int i = array.length-1; i > 0; i--) { for (int j = 0; J The above is the entire contents of this article, I hope to be able to understand the Java bubble so

Bubble Sort Method : A record of a small number of keywords is like a bubble stepping up, the key word larger record is like a stone sinking, each trip has one of the largest stones to the bottom. Algorithmic Nature: (The maximum is the key point, must be put to the last, so the cycle) each time from the first backward scrolling comparison, so that the maximum v

Solve the problem as follows:1. Seek the 1+2!+3!+...+20! and.2. Gets the maximum and minimum values in the array.3. Select sort, bubble sort for array.4. Find the location of a number in the array (binary lookup).binary Find: for the lookup of the sequential table, the array is stored in sequential elements, by the following method:Define the variable Low,mid,hi

number of the second comparison, and the third time, You just need to compare the numbers except the last two, and so on ... In other words, without a comparison, each trip less than once, to a certain extent, reduce the amount of the algorithm. In terms of time complexity: 1. If our data is in positive order, just take a trip to complete the sorting. The required number of comparisons and the number of records moved to a minimum, namely: cmin=n-1; Mmin=0 So, the best time complexity for

]; }} N--; System.out.println ("N=" +N); //Step:5 for(inti = 0; I ){ //swap If the previous number is greater than the last number if(A[i] > a[i+1]) {A[i]= A[i] ^ a[i+1]; A[i+1] = a[i+1] ^A[i]; A[i]= A[i] ^ a[i+1]; }} N--; returnA; }Add a For loop: Public Long[] Sort (Long[] a) { intn = a.length-1; for(intj = 0; J ){ for(inti = 0; I ){ //swap If the previous number is greater tha

1 Packagedemo;2 3 Importjava.util.Arrays;4 5 Public classDemo02 {6 Public Static voidMain (string[] args) {7 int[] arr =New int[10];8 for(inti=0; i){9Arr[i] = (int) (Math.random () *100);Ten } OneSystem.out.println ("Before bubbling Sort:"); ASystem.out.println ("No." + (0) + "Times:" +arrays.tostring (arr)); - - Bubblesort (arr); theSystem.out.println ("Before bubbling Sor