我正在尝试创建一个JavaScript函数,它将完整二进制文件的HTML返回到一定数量的位。
例如,full_bintree(1)
会为
提供HTML /\
1 0
和full_bintree(2)
会为
/\
/ \
1 0
/ \ / \
1 01 0
和full_bintree(3)
会为
/\
/ \
/ \
/ \
/ \
1 0
/ \ / \
/ \ / \
1 0 1 0
/ \ / \ / \ / \
1 0 1 01 0 1 0
等等。
我开始做一个功能,但它看起来很糟糕,我一直在意识到我需要回去修复的问题。
function full_bintree(var b)
{
var retstr = "";
var nr = (2 << b) + (b - 1); // number of rows
for (var i = 0, j = nr - 1, k = 0; i < nr; ++i, --j, k += 2)
{
retstr += "<p>";
// Add leading spaces on line
for (var m = 0; m < j; ++m)
retstr += " ";
// Add backslashes or
// Add middle spaces
for (var m = 0; m < k; ++m)
retstring += " ";
// Add forward slashes
retstr += "</p>";
}
}
答案 0 :(得分:2)
这是你的功能:
var logtree = function(A){
for(var i = 0; i < A.length; i++)
A[i] = A[i].join("");
A = A.join("\n");
if(typeof $ != "undefined")
$("#a").html(A);
console.log(A);
return A;
},
bintree = function(n, A, offset, lvl){
if(typeof offset == "undefined") offset = 0;
if(typeof lvl == "undefined") lvl = 0;
var width = 3 * Math.pow(2, n) - 1,
height = Math.ceil(width / 2),
mid = Math.floor(width / 2),
half = Math.ceil(height / 2),
nsub = Math.pow(2, lvl),
lim = offset + half - 1,
l = "/",
r = "\\";
if(typeof A == "undefined"){
A = [];
for(var i = 0; i < height; i++){
a = [];
for(var j = 0; j < width; j++) a.push(" ");
A[i] = a;
}
}
for(var i = offset, inc = 0; i <= lim; i++){
if(i == lim){
l = 1;
r = 0;
}
var a = 0, j = 0;
for(var j = 0; j < nsub; j++){
A[i][mid - inc + a - 1] = l;
A[i][mid + inc + a + 1] = r;
a += width + 1;
}
inc++;
}
if(n > 1)
bintree(n - 1, A, offset + height / 2, ++lvl);
return A;
};
logtree(bintree(4));