稀疏数组,实现,原理,应用;小Demo;学习笔记;

0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0	
0	0	0	0	0	0	0	0	0	0	0

假设上面就是一个11*11的五子棋,0代表空位,1 代表白棋,2代表黑黑旗;

其实有用的就是1,2;要是把0,1,2全存下,有点浪费,

这时候用稀疏数组,记录,可以减小空间;

稀疏数组的处理方法是:

1.记录数组有几行激烈,有多少个不同的值;

2.把具有不同值的行列记录在一个小规模的数组中,从而所以程序的规模;

稀疏数组的第一行存多少列,多少行,有多少个值;

接下来都是存值 的位置, 行,列,值;这种方式;

 

demo:

 

int chessArr[][] = new int[11][11];

        chessArr[1][2] = 1;
        chessArr[2][3] = 2;

        System.out.println("原数组是:");
        for (int[] row : chessArr) {
            for (int i : row) {
                System.out.printf("%d\t", i);
            }
            System.out.println();
        }

        //由于稀疏数组根据多少值来确定行数,先得到有多少值

        int sum = 0;
        for (int i = 0; i < chessArr.length; i++) {
            for (int j = 0; j < chessArr[i].length; j++) {
                if (chessArr[i][j] != 0) {
                    sum++;
                }
            }
        }
        System.out.println("总共有:" + sum + "个值");

        //接下来转成稀疏数组
        int sparseArr[][] = new int[sum + 1][3];

        //稀疏数组的第一行
        sparseArr[0][0] = 11;
        sparseArr[0][1] = 11;
        sparseArr[0][2] = sum;

        //存值
        int count = 0;
        for (int i = 0; i < chessArr.length; i++) {
            for (int j = 0; j < chessArr[i].length; j++) {
                if (chessArr[i][j] != 0) {
                    count++;
                    sparseArr[count][0] = i;
                    sparseArr[count][1] = j;
                    sparseArr[count][2] = chessArr[i][j];
                }
            }
        }

        //
        System.out.println("转换成稀疏数组");
        for (int[] row : sparseArr) {
            for (int i : row) {
                System.out.printf("%d\t", i);
            }
            System.out.println();
        }

        //然后将稀疏数组还原成二维数组

        int chessArr2[][] = new int[sparseArr[0][0]][sparseArr[0][1]];

      
        for (int i = 1; i < sparseArr.length; i++) {
            chessArr2[sparseArr[i][0]][sparseArr[i][1]] = sparseArr[i][2];
        }

        System.out.println("还原后的棋谱");

        for (int[] row : chessArr2) {
            for (int i : row) {
                System.out.printf("%d\t",i);
            }
            System.out.println();
        }

/./值:

原数组是:
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	
总共有:2个值
转换成稀疏数组
11	11	2	
1	2	1	
2	3	2	

还原后的棋谱
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