稀疏数组

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;
        //输出原始数组
        System.out.println("输出原始数组");

        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
        //获取有效值个数
        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("有效值个数:"+sum);

        //2.创建一个稀疏数组的数列
        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,将非零值存放在稀疏数组中
        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];

                }
            }
        }

        //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]+"\t");
        }

        //1.读取稀疏数组
        int[][] array3 = new int[array2[0][0]][array2[0][1]];

        //2.给其中的元素还原他的值
        for (int i = 1; i < array2.length; i++) {
            array3[array2[i][0]][array2[i][1]] = array2[i][2];
        }

        //3.打印
        System.out.println("输出还原数组");

        for (int[] ints : array3) {
            for (int anInt : ints) {
                System.out.print(anInt+"\t");
            }
            System.out.println();
        }

    }
posted @ 2021-05-26 11:57  做一个好人  阅读(33)  评论(0)    收藏  举报