29 java数组_8 _稀疏数组

java数组_8 _稀疏数组

介绍

1. 稀疏数组是一种数据结构

2. 当一个数组中大部分元素为0，或者为同一值的数组时，可以用稀疏数组来保存该数组

3. 稀疏数组的处理方式为：

   1. 记录数组一共有几行几列，有多少个不同值
   2. 把具有不同值的元素和行列及值记录在一个小规模的数组中，从而缩小程序的规模

4. 如下图：左边是原始数组，右边是稀疏数组

例子：（记录棋盘）

​

package com.wlw.arrays;

public class ArraysDemo08 {
    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[][] a1 = new int[11][11];
        a1[1][2] = 1;
        a1[2][3] = 2;

        System.out.println("输出原始数组：" );
/*        for (int[] ints : a1) {
            for (int anInt : ints) {
                System.out.print(anInt+" ");
            }
            System.out.println();
        }*/

        for (int i = 0; i < a1.length; i++) {
            for (int j = 0; j < a1[i].length; j++) {
                System.out.print(a1[i][j] + " ");
            }
            System.out.println();
        }

        System.out.println("================================");

        //转换为稀疏数组
        //1.先来计算原数组有几个非0数(有效值)
        int sum = 0;
        for (int i = 0; i < a1.length; i++) {
            for (int j = 0; j < a1[i].length; j++) {
                if (a1[i][j] != 0) {
                    sum++;
                }
            }
        }

        System.out.println("原数组中有效值的个数为："+sum);

        //2.创建稀疏数组,稀疏数组就3列是确定的
        int[][] a2 = new int[sum+1][3];
        a2[0][0] = a1.length;
        a2[0][1] = a1[0].length;
        a2[0][2] = sum;

        //3.存放原数组有效值的行列及值
        int count = 0;
        for (int i = 0; i < a1.length; i++) {
            for (int j = 0; j < a1[i].length; j++) {
                if(a1[i][j] != 0 ){
                    count++;
                    a2[count][0] = i;
                    a2[count][1] = j;
                    a2[count][2] = a1[i][j];
                }
            }
        }

        //4.输出稀疏数组
        System.out.println("稀疏数组：");
        for (int i = 0; i < a2.length; i++) {
            System.out.println(a2[i][0] + "\t" + a2[i][1] + "\t" + a2[i][2] + "\t");
        }
        /*for (int[] ints : a2) {
            for (int anInt : ints) {
                System.out.print(anInt+" ");
            }
            System.out.println();
        }*/

        System.out.println("================================");
        System.out.println("还原");

        //1.创建一个数组
        int[][] a3 = new int[a2[0][0]][a2[0][1]];

        //2.赋值
        for (int i = 1; i < a2.length; i++) {
            a3[a2[i][0]][a2[i][1]] = a2[i][2];
        }

        //3.输出
        for (int[] ints : a3) {
            for (int anInt : ints) {
                System.out.print(anInt + " ");
            }
            System.out.println();
        }

    }
}
/*输出结果：
    输出原始数组：
    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 
    ================================
    原数组中有效值的个数为：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 

*/