我和HTML一样：
<div id="tree">
<div id="AND" class="node">
<div id="OR" class="node">
<div id="a" class="node"></div>
<div id="b" class="node"></div>
</div>
<div id="OR" class="node">
<div id="c" class="node"></div>
<div id="d" class="node"></div>
</div>
</div>
</div>
我使用jquery来获取这样的节点：
$(".node")
我使用for来形成这样的数组：
var expresion = ["AND","OR","a", "b", "OR", "c", "d"];
我想构建类似的东西：
a OR b AND c OR d
有谁知道构建布尔表达式的算法。 提前谢谢。
答案 0 :(得分：1)
尝试遍历数组的前缀：您将能够构建所需的表达式。这对于程序员来说是一个简单的过程，它将作为练习留下。 我会伸出援助之手：阅读Tree Traversal。基本上，数组访问的顺序将产生你想要的输出。
答案 1 :(得分：0)
function travarse(node){
var result = [];
var op = node.children("#AND,#OR");
var len = op.length;
var term;
if(len){
term = op.children(".node");
result=result.concat(travarse($(term.get(0))));
result.push(op.get(0).id);
result=result.concat(travarse($(term.get(1))));
} else {
term = node.children(".node");
result.push(term.get(0).id);
result.push(node.get(0).id);
result.push(term.get(1).id);
}
return result;
}
//var expresion = $.map($.makeArray($(".node")), function(el){ return el.id; });//["AND", "OR", "a", "b", "OR", "c", "d"]
var expression = travarse($('#tree'));//["a", "OR", "b", "AND", "c", "OR", "d"]
答案 2 :(得分：0)
理想情况下，您应该只对节点进行自己的有序遍历。 请参阅：http://en.wikipedia.org/wiki/In-order_traversal#Inorder_Traversal
但是，此功能应该为您提供您正在寻找的确切功能：
function convertPreOrderArrayToInOrderString(elements)
{
var cursor = 0;
// While all the elements aren't merged together
while (elements.length > 1)
{
// Skip to the nextleaf
while (elements[cursor] === "AND" || elements[cursor] === "OR") cursor++;
// If we made all the matches we can at this depth, go to the next one
if (typeof elements[cursor+1] === "undefined")
{
cursor = 0;
continue;
}
// The relevant pieces to combining
var previous = elements[cursor-1];
var current = elements[cursor];
var next = elements[cursor+1];
// Create a string from the pieces
var tmpStr = current + " " + previous;
// How many we've combined
var combined = 2;
// If the third piece is complete, we can join it too
if (next !== "AND" && next != "OR")
{
// Tack the third piece on
tmpStr += " " + next;
combined++;
}
// Remove the individual elements and replace with the combined ones
var newElements = elements.slice(0, cursor-1);
newElements.push(tmpStr);
newElements = newElements.concat(elements.slice(cursor - 1 + combined));
// From now on, opreate on the new array with combined elements
elements = newElements;
}
// There should be one element left
return elements[0];
}
var expression = ["AND","OR","AND","a","f","OR","e","b","OR","c","d"];
console.log(convertPreOrderArrayToInOrderString(expression));
// Gives: "a AND f OR e OR b AND c OR d"
相应的小提琴：http://jsfiddle.net/DvZj4/4/
在更一般的情况下，您可能更喜欢这样做：
function convertPreOrderToInOrder(elements)
{
// Very simple sanity checks
if (elements.length === 1 && (elements[0] === "AND" || elements[0] === "OR"))
{
console.error("Cannot construct compound expression with boolean connectors without values");
return;
}
else if (elements.length === 1)
{
// Only 1 items, just return it
return elements[0];
}
else if (elements.length > 1 && elements[0] !== "AND" && elements[0] !== "OR")
{
console.error("Cannot construct compound expression without boolean connectors");
return;
}
// Convert back to a tree from the preorder, then flatten again in order
return flattenTreeInOrder(getTreeFromPreOrderArray(elements));
}
// Reconstructs a tree from a pre-ordered array of elements
// Assumes your array of elements can construct a properly
// balanced tree. If it can't, this will get screwy
function getTreeFromPreOrderArray(elements)
{
var i, el, root = {}, current = root;
// Iterate over every element
for (i in elements)
{
el = elements[i];
// Pretty much just a root check
if ( (el === "AND" || el === "OR") && !current.connector)
{
// Make this a root
current.connector = el;
}
// Not a root, can we go left?
else if (!current.left)
{
// Branch left
if (el === "AND" || el === "OR")
{
current.left = {connector: el};
current.left.parent = current;
current = current.left;
}
// Left left
else
{
current.left = el;
current.left.parent = current;
}
}
// Not a root, can't go left. Try to the right.
else
{
// Branch right
if (el === "AND" || el === "OR")
{
current.right = {connector: el};
current.right.parent = current;
current = current.right;
}
// Leaf Right
else
{
current.right = el;
current.right.parent = current;
// Tricky bit: If we put a leaf on the right, we need
// to go back up to the nearest parent who doesn't have
// right child;
while (current.parent && current.right)
{
current = current.parent;
}
}
}
}
return root;
}
// Flatten a tree into an array using in-order traversal
// No recursion, uses stack-based traversal
// Returns an array of strings
function flattenTreeInOrder(root)
{
var stack = [];
var flat = [];
var current = root;
// We need to keep going as long as there's either something in the
// stack, or we're looking at a node.
while (stack.length > 0 || current)
{
// If the looked at node is defined
// This will help us tell if we've gone past a leaf
if (current)
{
// Don't care during in order traversal. Just push this on
// and move to the left
stack.push(current);
current = current.left;
}
else
{
// Starting to go back up, but the current is undefined
// So, pop first, then start capturing in order elements
current = stack.pop();
flat.push(current.connector || current);
current = current.right;
}
}
return flat;
}
var expression = ["AND","OR","AND","a","f","OR","e","b","OR","c","d"];
console.log(convertPreOrderToInOrder(expression).join(" "));
// Gives: "a AND f OR e OR b AND c OR d"
相应的小提琴：http://jsfiddle.net/keR5j/
id s 此外，您不能以结构中的方式使用ID。拥有相同id的多个实例，即具有id="OR"的两个节点，每个规范都是无效的HTML：http://www.w3.org/TR/html401/struct/global.html#h-7.5.2
此属性为元素指定名称。该名称在文档中必须是唯一的。