In previous implement about the Segment Tree by Artur Antonov, it's a strong data structure with Generic <T>. And we can pass a closure parameter function: (T, T) -> T to reflect the relationship between parent and child node. And in particular, Generic can solve multiple strings stitching problem. It's just like the sample in Playground:
在之前关于Artur Antonov的线段树的实现中，它是一个强大的数据结构，带有泛型 <T>。 我们可以传递一个闭包参数function:(T, T) -> T来反映父节点和子节点之间的关系。 特别是，泛型可以解决多个字符串拼接问题。 它就像 Playground 中的样本：
stringSegmentTree.replaceItem(at: 0, withItem: "I")
stringSegmentTree.replaceItem(at: 1, withItem: " like")
stringSegmentTree.replaceItem(at: 2, withItem: " algorithms")
stringSegmentTree.replaceItem(at: 3, withItem: " and")
stringSegmentTree.replaceItem(at: 4, withItem: " swift")
stringSegmentTree.replaceItem(at: 5, withItem: "!")
print(stringSegmentTree.query(leftBound: 0, rightBound: 5))
// "I like algorithms and swift!"The use of <T> is so exciting. But we seldom use the Segment Tree to solve string problem instead of Suffix Array. And the Segment Tree is a kind of Interval Tree to solve the Interval Problem in mathemtics and statistics, which is a structure for storing intervals, or segments, and allows querying which of the stored segments contain a given point. A segment tree for a set I of n intervals uses O(nlogn) storage and can be built in O(nlogn) time. Segment trees support searching for all the intervals that contain a query point in O(log n+k), k being the number of retrieved intervals or segments.
使用<T>是如此令人兴奋。 但是我们很少使用线段树来解决字符串问题而不是 Suffix Array。 并且线段树是一种Interval Tree来解决mathemtics和statistics中的Interval Problem，它是一种存储间隔或段的结构，并允许查询哪些存储的段包含给定的点。 n个区间的集合I的分段树使用O(nlogn)存储，并且可以在O(nlogn)时间内构建。 段树支持搜索包含O(log n+k)中的查询点的所有间隔，k是检索的间隔或段的数量。
But that is common Segment Tree. By Lazy Propagation, we can implement to modify an interval in O(logn) time. Let's explore together in following:
但这是常见的细分树。 通过Lazy Propagation，我们可以实现在O(logn)时间内修改间隔。 让我们一起探讨如下：
At first, we reference the implement of Artur Antonov about Segment Tree. This code contained build, single update and interval query three operation. The implement of build and single update operation is following: 首先，我们参考Artur Antonov关于线段树的实现。 此代码包含build，single update和interval query三个操作。 build和single update操作的实现如下：
// Author: Artur Antonov
public init(array: [T], leftBound: Int, rightBound: Int, function: @escaping (T, T) -> T) {
self.leftBound = leftBound
self.rightBound = rightBound
self.function = function
// ①
if leftBound == rightBound {
value = array[leftBound]
}
// ②
else {
let middle = (leftBound + rightBound) / 2
leftChild = SegmentTree<T>(array: array, leftBound: leftBound, rightBound: middle, function: function)
rightChild = SegmentTree<T>(array: array, leftBound: middle+1, rightBound: rightBound, function: function)
value = function(leftChild!.value, rightChild!.value)
}
}In position ①, it means the current node is leaf because its left bound data is equal to the right. So we assign a value to it directly. In position ②, it means the current node is parent (which has one or more children), and we need to recursion down and update current node's data in the follow-up process. 在位置①，它表示当前节点是leaf，因为它的左边界数据等于右边。 所以我们直接给它赋值。 在位置②，它表示当前节点是parent（有一个或多个子节点），我们需要递归并在后续过程中更新当前节点的数据。
And then, we have a look for interval query operation: 然后，我们来看看interval query操作：
// Author: Artur Antonov
public func query(leftBound: Int, rightBound: Int) -> T {
if self.leftBound == leftBound && self.rightBound == rightBound {
return self.value
}
guard let leftChild = leftChild else { fatalError("leftChild should not be nil") }
guard let rightChild = rightChild else { fatalError("rightChild should not be nil") }
// ①
if leftChild.rightBound < leftBound {
return rightChild.query(leftBound: leftBound, rightBound: rightBound)
}
// ②
else if rightChild.leftBound > rightBound {
return leftChild.query(leftBound: leftBound, rightBound: rightBound)
}
// ③
else {
let leftResult = leftChild.query(leftBound: leftBound, rightBound: leftChild.rightBound)
let rightResult = rightChild.query(leftBound:rightChild.leftBound, rightBound: rightBound)
return function(leftResult, rightResult)
}
}Position ① means that the left bound of current query interval is on the right of this right bound, so recurs to right direction. Position ② is opposite of position ①, recurs to left direction. Position ③ means our check interval is included the interval we need, so recurs deeply. 位置①表示当前查询间隔的左边界位于此右边界的右侧，因此重复到右边方向。 位置②与位置①相反，向左方向重复。 位置③表示我们的检查间隔包含在我们需要的间隔内，因此深度重复。
There are common part from the two parts of code above - recurs deeply below, and update data up. So we can decouple this operation named func pushUp(lson: LazySegmentTree, rson: LazySegmentTree):
上面代码的两个部分有共同的部分 - 在下面深入重复，并且更新数据。 所以我们可以解耦这个名为func pushUp(lson: LazySegmentTree, rson: LazySegmentTree)的操作：
// MARK: - Push Up Operation
private func pushUp(lson: LazySegmentTree, rson: LazySegmentTree) {
self.value = lson.value + rson.value
}(This code only describe the Sum Segment Tree)
And then, we can update the implement before:
public init(array: [Int], leftBound: Int, rightBound: Int) {
self.leftBound = leftBound
self.rightBound = rightBound
self.value = 0
self.lazyValue = 0
guard leftBound != rightBound else {
value = array[leftBound]
return
}
let middle = leftBound + (rightBound - leftBound) / 2
leftChild = LazySegmentTree(array: array, leftBound: leftBound, rightBound: middle)
rightChild = LazySegmentTree(array: array, leftBound: middle + 1, rightBound: rightBound)
if let leftChild = leftChild, let rightChild = rightChild {
pushUp(lson: leftChild, rson: rightChild)
}
}
// MARK: - One Item Update
public func update(index: Int, incremental: Int) {
guard self.leftBound != self.rightBound else {
self.value += incremental
return
}
guard let leftChild = leftChild else { fatalError("leftChild should not be nil") }
guard let rightChild = rightChild else { fatalError("rightChild should not be nil") }
let middle = self.leftBound + (self.rightBound - self.leftBound) / 2
if index <= middle { leftChild.update(index: index, incremental: incremental) }
else { rightChild.update(index: index, incremental: incremental) }
pushUp(lson: leftChild, rson: rightChild)
}You may feel that the pushUp is so simple. In fact, this's just to lead pushDown this function.
Before this, I want to talk about the topic about interval operation. The interval operation is a way to update all elements of a continuous subset. But these isn't in the version of Artur Antonov. You might disdain with this, because it's solved with a for loop:
您可能会觉得pushUp非常简单。 实际上，这只是为了引导pushDown这个功能。
在此之前，我想谈谈关于间隔操作的话题。 间隔操作是更新连续子集的所有元素的一种方法。 但这些不是Artur Antonov的版本。 你可能会不屑于此，因为它是通过for循环解决的：
// Sample: update the elements with subscript [2, 5]
for index in 2 ... 5 {
sumSegmentTree.update(index: index, incremental: 3)
}It is a O(n) time operation, which make the interval operation uses O(nlogn) time to update these elements. We need a O(logn) way to maintain the elegance of Segment Tree.
Check the data structure of Segment Tree again:
它是一个O(n)时间操作，它使间隔操作使用O(nlogn)时间来更新这些元素。 我们需要一种O(logn)方式来保持线段树的优雅。
再次检查Segment Tree的数据结构：
We only catch the root node in programming. If we want to explore the bottom of the tree, and use pushUp to update every node, the task will be reached. So it asked us to traverse the tree, that spent O(n) time to do this with any way. This can't conform our expectations.
我们只在编程中捕获根节点。 如果我们想要探索树的底部，并使用pushUp来更新每个节点，那么将会到达任务。 因此，它要求我们遍历树，花费O(n)的时间以任何方式做到这一点。 这不符合我们的期望。
Then we started to think about pushDown to update down from the root. After we update the parent, the data continued to distributed to its children according to law. But it still need O(n) time to do this. Keep thinking, we only update the parent, and to update the children when query time. Yeah, that's the key of lazy propagation. Because the recursing direct of the query and update interval is same. So we got it! 😁 Let's check this sample:
然后我们开始考虑pushDown从根更新。 在我们更新父母之后，数据继续依法分发给孩子们。但是仍然需要O(n)时间来做这件事。 继续思考，我们只更新父级，并在query time 时更新子级。 是的，这是懒惰传播的关键。 因为query和update interval的递归直接相同。 所以我们明白了！ 😁我们来看看这个样本：
update make the subscript 1...3 elements plus 2, so we make the 1st node in 2 depth and 3rd in 3 depth get a lazy mark, which means these node need to be updated. And we shouldn't add a lazy mark for root node, because it was updated before the pushDown in the first recursing.
update使下标 1...3元素加2，所以我们使第1个节点在2深度和3在3深度得到懒惰标记，这意味着这些节点需要更新。 我们不应该为根节点添加lazy mark，因为它在第一次递归中的pushDown之前更新了。
In query operation, we accord to the original method to recurs the tree, and find the 1st node held lazy mark in 2 depth, so to update it. It's the same situation about the 1st node in 3 depth.
在query操作中，我们按照原始方法重新生成树，并在2深度找到第一个节点保持lazy mark，以便更新它。 关于3深度的第1个节点，情况也是如此。
Do you understand the lazy propagation? In short, we only update the wide range node data and add it a lazy mark. Then they will be update when we need to query them. And the Update Down operation is the function of pushDown.
你了解懒惰传播？ **简而言之，我们只更新宽范围节点数据并添加一个懒惰标记。 然后当我们需要查询它们时它们会更新。**并且Update Down操作是pushDown的功能。
This is the complete implementation about the Sum Segment Tree with interval update operation: 这是关于具有间隔更新操作的Sum Segment Tree的完整实现：
public class LazySegmentTree {
private var value: Int
private var leftBound: Int
private var rightBound: Int
private var leftChild: LazySegmentTree?
private var rightChild: LazySegmentTree?
// Interval Update Lazy Element
private var lazyValue: Int
// MARK: - Push Up Operation
// Description: pushUp() - update items to the top
private func pushUp(lson: LazySegmentTree, rson: LazySegmentTree) {
self.value = lson.value + rson.value
}
// MARK: - Push Down Operation
// Description: pushDown() - update items to the bottom
private func pushDown(round: Int, lson: LazySegmentTree, rson: LazySegmentTree) {
guard lazyValue != 0 else { return }
lson.lazyValue += lazyValue
rson.lazyValue += lazyValue
lson.value += lazyValue * (round - (round >> 1))
rson.value += lazyValue * (round >> 1)
lazyValue = 0
}
public init(array: [Int], leftBound: Int, rightBound: Int) {
self.leftBound = leftBound
self.rightBound = rightBound
self.value = 0
self.lazyValue = 0
guard leftBound != rightBound else {
value = array[leftBound]
return
}
let middle = leftBound + (rightBound - leftBound) / 2
leftChild = LazySegmentTree(array: array, leftBound: leftBound, rightBound: middle)
rightChild = LazySegmentTree(array: array, leftBound: middle + 1, rightBound: rightBound)
if let leftChild = leftChild, let rightChild = rightChild {
pushUp(lson: leftChild, rson: rightChild)
}
}
public convenience init(array: [Int]) {
self.init(array: array, leftBound: 0, rightBound: array.count - 1)
}
public func query(leftBound: Int, rightBound: Int) -> Int {
if leftBound <= self.leftBound && self.rightBound <= rightBound {
return value
}
guard let leftChild = leftChild else { fatalError("leftChild should not be nil") }
guard let rightChild = rightChild else { fatalError("rightChild should not be nil") }
pushDown(round: self.rightBound - self.leftBound + 1, lson: leftChild, rson: rightChild)
let middle = self.leftBound + (self.rightBound - self.leftBound) / 2
var result = 0
if leftBound <= middle { result += leftChild.query(leftBound: leftBound, rightBound: rightBound) }
if rightBound > middle { result += rightChild.query(leftBound: leftBound, rightBound: rightBound) }
return result
}
// MARK: - One Item Update
public func update(index: Int, incremental: Int) {
guard self.leftBound != self.rightBound else {
self.value += incremental
return
}
guard let leftChild = leftChild else { fatalError("leftChild should not be nil") }
guard let rightChild = rightChild else { fatalError("rightChild should not be nil") }
let middle = self.leftBound + (self.rightBound - self.leftBound) / 2
if index <= middle { leftChild.update(index: index, incremental: incremental) }
else { rightChild.update(index: index, incremental: incremental) }
pushUp(lson: leftChild, rson: rightChild)
}
// MARK: - Interval Item Update
public func update(leftBound: Int, rightBound: Int, incremental: Int) {
if leftBound <= self.leftBound && self.rightBound <= rightBound {
self.lazyValue += incremental
self.value += incremental * (self.rightBound - self.leftBound + 1)
return
}
guard let leftChild = leftChild else { fatalError("leftChild should not be nil") }
guard let rightChild = rightChild else { fatalError("rightChild should not be nil") }
pushDown(round: self.rightBound - self.leftBound + 1, lson: leftChild, rson: rightChild)
let middle = self.leftBound + (self.rightBound - self.leftBound) / 2
if leftBound <= middle { leftChild.update(leftBound: leftBound, rightBound: rightBound, incremental: incremental) }
if middle < rightBound { rightChild.update(leftBound: leftBound, rightBound: rightBound, incremental: incremental) }
pushUp(lson: leftChild, rson: rightChild)
}
}Explain some sample snippets:
private var lazyValue: IntHere we add a new property for Segment Tree to represent lazy mark. And it is a incremental value for Sum Segment Tree. If the lazyValue isn't equal to zero, the current node need to be updated. And its real value is equal to value + lazyValue * (rightBound - leftBound + 1).
这里我们为线段树添加一个新属性来表示lazy mark。 它是Sum Segment Tree的增量值。 如果lazyValue不等于零，则需要更新当前节点。 它的实际值等于 value + lazyValue * (rightBound - leftBound + 1)。
// MARK: - Push Down Operation
// Description: pushDown() - update items to the bottom
private func pushDown(round: Int, lson: LazySegmentTree, rson: LazySegmentTree) {
guard lazyValue != 0 else { return }
lson.lazyValue += lazyValue
rson.lazyValue += lazyValue
lson.value += lazyValue * (round - (round >> 1))
rson.value += lazyValue * (round >> 1)
lazyValue = 0
}
At first we check whether the node needs to be updated. If the lazyValue isn't equal to zero, we need to pushDown. And the update rules are the lazyValue * (rightBound - leftBound + 1). At last, reset the lazyValue to zero.
首先，我们检查节点是否需要更新。 如果lazyValue不等于零，我们需要pushDown。 更新规则是 lazyValue * (rightBound - leftBound + 1)。 最后，将lazyValue重置为零。
This is the introduce of the Lazy Propagation in Segment Tree. You can also learn Functional Segment Tree to understand the Lazy Propagation deeply. In addition, I learn from the notonlysuccess's code about Segment Tree described by C, this is the link. 这是在线段树中引入懒惰的传播。 您还可以学习 Functional Segment Tree 以深入理解Lazy Propagation。 另外，我从C语言描述的关于Segment Tree的 notonlysuccess 代码中学习，这是link。
In fact, the operation of Segment Tree is far more than that. It can also be used to handle problems between collections. I want to implement it with Swift in the future and make the Swift Segment Tree stronger and stronger. 😁 实际上，线段树的操作远不止于此。 它还可以用于处理集合之间的问题。 我想在未来使用Swift实现它，并使Swift Segment Tree更强大。😁
Written for Swift Algorithm Club by Desgard_Duan
作者：Desgard_Duan
翻译：Andy Ron