對(duì)一個(gè)數(shù)組進(jìn)行zig-zag重新排列

2020-01-26 16:21:16

字體：大中小

供稿：網(wǎng)友

在看jpeg解碼，里面有對(duì)8x8數(shù)組進(jìn)行重排。里面直接提供了unzig表：

int unzig[] = {

0, 1, 8, 16, 9, 2, 3, 10,
17, 24, 32, 25, 18, 11, 4, 5,
12, 19, 26, 33, 40, 48, 41, 34,
27, 20, 13, 6, 7, 14, 21, 28,
35, 42, 49, 56, 57, 50, 43, 36,
29, 22, 15, 23, 30, 37, 44, 51,
58, 59, 52, 45, 38, 31, 39, 46,
53, 60, 61, 54, 47, 55, 62, 63,

};

然后自己寫(xiě)了個(gè)產(chǎn)生unzig表的程序。

:-)

代碼如下：

復(fù)制代碼代碼如下:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef void (*fn)(int, int, int, int, void*);

printpos(int x, int y, int n, int i, void *arr)
{
// printf("%2d%c", x+y*n, i%n==(n-1)?'/n':' ');
int *a;

a = (int*)arr;
printf("%2d%c", a[i], i%n==(n-1)?'/n':' ');
}

unzigasgn(int x, int y, int n, int i, void *arr)
{
int *a;

a = (int*)arr;
a[i] = x+y*n;
}

zigasgn(int x, int y, int n, int i, void *arr)
{
int *a;

a = (int*)arr;
a[x+y*n] = i;
}

zigzag(int n, fn f1, void *arr)
{
int i, x, y;

    i = 0;
    x = y = 0;
    f1(x, y, n, i, arr);
    for(;;) {
        /* right, or down */
        if(++i >= n*n)
            return;
        if(x+1 < n){
            x++;
            f1(x, y, n, i, arr);
        }else{
            y++;
            f1(x, y, n, i, arr);
        }

        /* left down */
        while(x-1 >= 0 && y+1 < n){
            x--;
            y++;
            if(++i >= n*n)
                return;
            f1(x, y, n, i, arr);
        }

        /* down, or right */
        if(++i >= n*n)
            return;
        if(y+1 < n){
            y++;
            f1(x, y, n, i, arr);
        }else{
            x++;
            f1(x, y, n, i, arr);
        }

        /* right up */
        while(x+1 < n && y-1 >= 0){
            x++;
            y--;
            if(++i >= n*n)
                return;
            f1(x, y, n, i, arr);
        }
    }
}

testzigzag(int n)
{
    int i;
    int n2;
    int *arr;

n2 = n*n;
arr = malloc(n*n*sizeof(*arr));

    zigzag(n, (fn)zigasgn, arr);
    zigzag(n, (fn)printpos, arr);
    printf("/n/n");
    zigzag(n, (fn)unzigasgn, arr);
    zigzag(n, (fn)printpos, arr);
}

main(int argc, char **argv)
{
int n;

    n = 8;
    if(argc > 1){
        n = atoi(argv[1]);
    }
    testzigzag(n);
    return 0;
}

運(yùn)行結(jié)果截圖：