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@justinbowes
Created August 8, 2012 19:11
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JavaScript binary heap with merge support
var BinaryHeap = function BinaryHeap(scoreFunction, populate){
this.content = (populate ? populate.slice(0) : []);
this.scoreFunction = scoreFunction;
if (this.content.length) {
this.build();
}
}
BinaryHeap.prototype = {
/**
* Create a heap from a random arrangement of data.
* Building a heap this way is O(n).
*/
build: function() {
var lastNonLeaf = Math.floor(this.content.length / 2) - 1;
for (var i = lastNonLeaf; i >= 0; i--) {
this.sinkDown(i);
}
},
/**
* Adds an element to the heap.
* Building a heap this way is O(n log n).
*/
push: function(element) {
// Add the new element to the end of the array.
this.content.push(element);
// Allow it to bubble up.
this.bubbleUp(this.content.length - 1);
},
/**
* Pops the root off the heap. The root is the element for which
* the score function returns the minimum score. This is delete-min
* with O(log n).
*/
pop: function() {
// Store the first element so we can return it later.
var result = this.content[0];
// Get the element at the end of the array.
var end = this.content.pop();
// If there are any elements left, put the end element at the
// start, and let it sink down.
if (this.content.length > 0) {
this.content[0] = end;
this.sinkDown(0);
}
return result;
},
/**
* Inspects the root of the heap. This is find-min, O(1).
*/
peek: function() {
return (this.content && this.content.length ? this.content[0] : null);
},
/**
* Removes the given element from this heap.
*/
remove: function(element) {
var len = this.content.length;
// To remove a value, we must search through the array to find
// it.
for (var i = 0; i < len; i++) {
if (this.content[i] == element) {
// When it is found, the process seen in 'pop' is repeated
// to fill up the hole.
var end = this.content.pop();
if (i != len - 1) {
this.content[i] = end;
if (this.scoreFunction(end) < this.scoreFunction(node)) this.bubbleUp(i);
else this.sinkDown(i);
}
return;
}
}
throw new Error("Node not found.");
},
/**
* Gets the heap size. I assume JavaScript allows O(n) for this.
*/
size: function() {
return this.content.length;
},
/**
* Bubbles up an element at a given index. This operation is O(log n).
*/
bubbleUp: function(n) {
// Fetch the element that has to be moved.
var element = this.content[n];
// When at 0, an element can not go up any further.
while (n > 0) {
// Compute the parent element's index, and fetch it.
var parentN = Math.floor((n + 1) / 2) - 1,
parent = this.content[parentN];
if (this.scoreFunction(element) < this.scoreFunction(parent)) {
// Swap the elements if the parent is greater.
this.content[parentN] = element;
this.content[n] = parent;
// Update 'n' to continue at the new position.
n = parentN;
} else {
// Found a parent that is less, no need to move it further.
break;
}
}
},
/**
* Sinks down an element at a given index. This operation is O(log n).
*/
sinkDown: function(n) {
// Look up the target element and its score.
var length = this.content.length,
element = this.content[n],
elemScore = this.scoreFunction(element);
do {
// Compute the indices of the child elements.
var child2N = (n + 1) * 2, child1N = child2N - 1;
// This is used to store the new position of the element,
// if any.
var swap = null;
// If the first child exists (is inside the array)...
if (child1N < length) {
// Look it up and compute its score.
var child1 = this.content[child1N],
child1Score = this.scoreFunction(child1);
// If the score is less than our element's, we need to swap.
if (child1Score < elemScore)
swap = child1N;
}
// Do the same checks for the other child.
if (child2N < length) {
var child2 = this.content[child2N],
child2Score = this.scoreFunction(child2);
if (child2Score < (swap == null ? elemScore : child1Score))
swap = child2N;
}
// If the element needs to be moved, swap it, and continue.
if (swap != null) {
this.content[n] = this.content[swap];
this.content[swap] = element;
n = swap;
}
} while (swap != null);
}
};
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