数字矩阵算法

 1 var canvas = document.getElementById('canvas'), ctx = canvas.getContext('2d');
 2     var start = 1, end = 256, step = 30, direction = 'right';
 3     var len = end - start + 1, wLen = 1, hLen = 1, flag = 1;
 4     var sideLength = Math.ceil(Math.sqrt(len)); // 总数求边长
 5     var currentX = step*(sideLength-1)/2, currentY = step*(sideLength+2)/2;
 6 
 7     ctx.moveTo(currentX, currentY);
 8     ctx.font = "14px sans-serif";
 9     // ctx.fillRect(0,0, canvas.width, canvas.height)
10     ctx.fillStyle = "#000";
11     console.log(sideLength);
12     while (len--) {
13         currentX += getXY(direction).x;
14         currentY += getXY(direction).y;
15         
16         ctx.fillText(start++, currentX, currentY);
17 
18         ctx.moveTo(currentX, currentY);
19         
20         if (direction == 'up' && flag > hLen) {
21             hLen++;
22             direction = 'right';
23             flag = 1;
24         } else if (direction == 'right' && flag > wLen) {
25             wLen++;
26             direction = 'down';
27             flag = 1;
28         } else if (direction == 'down' && flag > hLen) {
29             hLen++;
30             direction = 'left';
31             flag = 1;
32         } else if (direction == 'left' && flag > wLen) {
33             wLen++;
34             direction = 'up';
35             flag = 1;
36         }
37 
38         flag ++;
39     }
40 
41     function getXY(dir) {
42         var x = 0, y = 0;
43         switch(dir) {
44             case 'up':  
45                 x = 0;
46                 y -= step;
47                 break;
48             case 'right':  
49                 x += step;
50                 y = 0;
51                 break;
52             case 'down':  
53                 x = 0;
54                 y += step;
55                 break;
56             case 'left':  
57                 x -= step;
58                 y = 0;
59                 break;
60         }       
61         return {x: x, y: y}
62     }
63   

 

posted @ 2013-10-31 15:40  风之约  阅读(434)  评论(0编辑  收藏  举报