package com.kuang.array; public class ArrayDemo08 { public static void main(String[] args) { //1.创建一个二维数组 11*11 0:没有棋子 1:黑棋 2:白棋 int[][] array1 = new int[11][11]; array1[1][2] = 1; array1[2][3] =2; //输出原始的数组 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(); } System.out.println("================="); //转换为稀疏数组保存 //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]; array2[0][0]=11; array2[0][1]=11; array2[0][2]=sum; //遍历二维数组,将非零的值,存在放在稀疏数组中 int count=0; for (int i = 0; i <array1.length ; i++) { for (int j = 0; j < array1[i].length; j++) { if(array1[i][j]!=0){ count++; array2[count][0]=i; array2[count][1]=j; array2[count][2]=array1[i][j]; } } } 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]+"\t"); } System.out.println("================"); System.out.println("还原"); //1.读取稀疏数组 //int[][] array3 = new int[11][11]; int[][] array3 = new int[array2[0][0]][array2[0][1]]; for (int i = 1; i < array2.length; i++) { array3[array2[i][0]][array2[i][1]]=array2[i][2]; } //3.打印 System.out.println("还原原始的数组"); for (int[] ints : array3) { for (int anInt : ints) { System.out.print(anInt+"\t"); } System.out.println(); } } }