稀疏数组
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();
}
}
}


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