冒泡排序,稀疏数列结构

冒泡排序
package com.zjm.array;
//冒泡排序
//1.比较数组中两个相邻的元素,如果第一个数比第二个数大,我就交换他们的位置
//2.每一次比较,就会产生出一个最大或者最小的数字
//3.下一轮则可以少一次排序
//4.依次循环,直到结束

import java.util.Arrays;

public class bubbleSort {
public static void main(String[] args) {
int[] a = {1,23,4,6,7,2123,642,23,547,41};
int[] sort = sort(a);
System.out.println(Arrays.toString(sort));
}

public static int[] sort(int[] array){
//临时变量
int temp = 0;
//外层循环,判断我们要走多少次
for (int i = 0; i <array.length ; i++) {

boolean flag = false; //通过flag标识位减少没有意义的比较

//内层循环,比较两个数大小,如果第一个数比第二个数大,则交换位置
for (int j = 0; j <array.length-1-i ; j++) {
if(array[j+1]<array[j]){
temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
flag = true;
}
if (flag==false){
break;
}
}
}
return array;
}
}

稀疏数列结构

 

 

 

 

package com.zjm.array;

public class SparseArray {
public static void main(String[] args) {
//1.创建一个二维数组作为棋盘 11*11 0:没有棋子 1:黑色棋子 2:白色棋子
int[][] array1 = new int[11][11];
array1[1][2] = 1;
array1[2][3] = 2;
//输出原始数组
System.out.println("输出原始的数组:");
for (int[] ints:array1) {
for (int anInt : ints) {
System.out.print(anInt+"\t");
}
System.out.println();
}

//转换为稀疏数组保存
//1.获取有效值的个数
int sum = 0;
for (int i = 0; i < 11; i++) {
for (int j = 0; j < 11; j++) {
if (array1[i][j]!=0){
sum++;
}
}
}
System.out.println("有效值的个数:"+sum);
//2.创建一个稀疏数组的数组
int[][] array2 = new int[sum+1][3];
//数组头
array2[0][0] = 11;
array2[0][1] = 11;
array2[0][2] = sum;
//遍历二维数组,将非零值,存放到稀疏数组中
int count=0;
for (int i = 0; i < array1.length; i++) {
for (int j = 0; j < array1[1].length; j++) {
if (array1[i][j]!=0){
count++;
array2[count][0] = i;
array2[count][1] = j;
array2[count][2] = array1[i][j];
}
}
}
//输出稀疏数组
System.out.println("稀疏数组:");

for (int i = 0; i < array2.length; i++) {
System.out.println(array2[i][0]+"\t"+array2[i][1]+"\t"+array2[i][2]);
}

//还原稀疏数组
System.out.println("还原:");
//1.读取稀疏数组
int[][] array3 = new int[array2[0][0]][array2[0][1]];
//2.还原有效值
for (int i = 1; i < array2.length; i++) {
array3[array2[i][0]][array2[i][1]] = array2[i][2];
}
//3.打印
for (int[] ints : array3) {
for (int anInt : ints) {
System.out.print(anInt+"\t");
}
System.out.println();
}
}
}


输出:

输出原始的数组:
0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
有效值的个数:2
稀疏数组:
11 11 2
1 2 1
2 3 2
还原:
0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0

posted @ 2022-09-23 20:49  去公司搞点薯条  阅读(37)  评论(0)    收藏  举报