線段樹是一種特殊的樹結構。這種數據結構主要用於解決“線段”或者是“區間”這種特殊的數據,是算法競賽中的常客。在這一章,我們將從底層實現屬於我們自己的線段樹,完成線段樹的創建,查詢,更新三個操作,並且通過實際比較,看到線段樹解決“線段”相關問題的巨大優勢。
9-1 什麼是線段樹(區間樹)
- 對於有一類問題,我們關心的是線段(或者區間)
- 最經典的線段樹問題:區間染色
- 另一類經典問題:區間查詢
9-2 線段樹基礎表示
public class SegmentTree<E> {
private E[] tree;
private E[] data;
public SegmentTree(E[] arr){
data = (E[]) new Object[arr.length];
for (int i=0;i<arr.length;i++)
data[i] = arr[i];
tree = (E[]) new Object[4 * arr.length];
}
public int getSize(){
return data.length;
}
public E get(int index){
if (index<0 || index>=data.length)
throw new IllegalArgumentException("Index is illegal.");
return data[index];
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的左孩子節點的索引
private int leftChild(int index){
return 2*index + 1;
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的右孩子節點的索引
private int rightChild(int index){
return 2*index + 2;
}
}
9-3 創建線段樹
public class SegmentTree<E> {
private E[] tree;
private E[] data;
private Merger<E> merger;
public SegmentTree(E[] arr, Merger<E> merger){
this.merger = merger;
data = (E[]) new Object[arr.length];
for (int i=0;i<arr.length;i++)
data[i] = arr[i];
tree = (E[]) new Object[4 * arr.length];
buildSegmentTree(0,0,data.length-1);
}
//在treeIndex的位置創建表示區間[1....r]的線段樹
private void buildSegmentTree(int treeIndex,int l,int r){
if (l==r){
tree[treeIndex] = data[l];
return;
}
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
//int mid = (l+r)/2 爲避免l和r都特別大的時候使得l+r產生整形溢出的情況
int mid = l + (r-l)/2;
buildSegmentTree(leftTreeIndex,l,mid);
buildSegmentTree(rightTreeIndex,mid+1,r);
tree[treeIndex] = merger.merge(tree[leftTreeIndex],tree[rightTreeIndex]);
}
public int getSize(){
return data.length;
}
public E get(int index){
if (index<0 || index>=data.length)
throw new IllegalArgumentException("Index is illegal.");
return data[index];
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的左孩子節點的索引
private int leftChild(int index){
return 2*index + 1;
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的右孩子節點的索引
private int rightChild(int index){
return 2*index + 2;
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
res.append('[');
for (int i=0;i<tree.length;i++){
if (tree[i]!=null)
res.append(tree[i]);
else
res.append("null");
if (i!=tree.length-1)
res.append(", ");
}
res.append(']');
return res.toString();
}
}
public interface Merger<E> {
//這裏用戶定義兩個區間如何融合的規則
E merge(E a,E b);
}
現在測試一下:
public class Main {
public static void main(String[] args) {
Integer[] nums = {-2,0,3,-5,2,-1};
// SegmentTree<Integer> segTree = new SegmentTree<>(nums, new Merger<Integer>() {
// @Override
// public Integer merge(Integer a, Integer b) {
// return a+b;
// }
// });
SegmentTree<Integer> segTree = new SegmentTree<>(nums,(a,b)->a+b);
System.out.println(segTree);
}
}
9-4 線段樹中的區間查詢
//返回區間[queryL,queryR]的值
public E query(int queryL, int queryR){
if (queryL<0 || queryL>=data.length || queryR<0 || queryR>=data.length || queryL>queryR)
throw new IllegalArgumentException("Index is illegal");
return query(0,0,data.length-1,queryL,queryR);
}
//在以treeIndex爲根的線段樹中[l,,,,r]的範圍裏,搜索區間[queryL....qeuryR]的值
private E query(int treeIndex,int l,int r,int queryL,int queryR){
if ( l==queryL && r == queryR)
return tree[treeIndex];
int mid = l+(r-l)/2;
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
if (queryL>=mid+1)
return query(rightTreeIndex,mid+1,r,queryL,queryR);
else if (queryR <= mid)
return query(leftTreeIndex,l,mid,queryL,queryR);
E leftResult = query(leftTreeIndex,l,mid,queryL,mid);
E rightResult = query(rightTreeIndex,mid+1,r,mid+1,queryR);
return merger.merge(leftResult,rightResult);
}
我們測試一下:
public class Main {
public static void main(String[] args) {
Integer[] nums = {-2,0,3,-5,2,-1};;
SegmentTree<Integer> segTree = new SegmentTree<>(nums,(a,b)->a+b);
System.out.println(segTree.query(0,2));
System.out.println(segTree.query(2,5));
System.out.println(segTree.query(0,5));
}
}
9-5 Leetcode上線段樹相關的問題
public class NumArray {
private SegmentTree<Integer> segmentTree;
public NumArray(int[] nums){
if (nums.length>0){
Integer[] data = new Integer[nums.length];
for (int i=0;i<nums.length;i++)
data[i] = nums[i];
segmentTree = new SegmentTree<>(data,(a,b) -> a+b);
}
}
public int sumRange(int i,int j){
if (segmentTree == null)
throw new IllegalArgumentException("Segment Tress is null");
return segmentTree.query(i,j);
}
}
想快速查詢某個區間的元素和,而且這個區間中的元素不會改變,對於這樣的需求可以進行預處理
public class NumArray {
private int[] sum; // sum[i]存儲前i個元素和, sum[0] = 0
// 即sum[i]存儲nums[0...i-1]的和
// sum(i, j) = sum[j + 1] - sum[i]
public NumArray(int[] nums) {
sum = new int[nums.length + 1];
sum[0] = 0;
for(int i = 1 ; i < sum.length ; i ++)
sum[i] = sum[i - 1] + nums[i - 1];
}
public int sumRange(int i, int j) {
return sum[j + 1] - sum[i];
}
}
/// Leetcode 307. Range Sum Query - Mutable
/// https://leetcode.com/problems/range-sum-query-mutable/description/
///
/// 使用sum數組的思路:TLE
class NumArray3 {
private int[] data;
private int[] sum;
public NumArray3(int[] nums) {
data = new int[nums.length];
for(int i = 0 ; i < nums.length ; i ++)
data[i] = nums[i];
sum = new int[nums.length + 1];
sum[0] = 0;
for(int i = 1 ; i <= nums.length ; i ++)
sum[i] = sum[i - 1] + nums[i - 1];
}
public int sumRange(int i, int j) {
return sum[j + 1] - sum[i];
}
public void update(int index, int val) {
data[index] = val;
for(int i = index + 1 ; i < sum.length ; i ++)
sum[i] = sum[i - 1] + data[i - 1];
}
}
使用數組雖然可以實現,但是耗時非常多,會超出時間限制。
這個時候就要用到我們的線段樹。
9-6 線段樹中的更新操作
//將index位置的值,更新爲e
public void set(int index,E e){
if (index<0 || index >= data.length)
throw new IllegalArgumentException("Index is illegal");
data[index] = e;
set(0,0,data.length-1,index,e);
}
//在以treeIndex爲根的線段樹中更新index的值爲e
private void set(int treeIndex,int l,int r,int index,E e){
if (l==r){
tree[treeIndex] = e;
return;
}
int mid = l+ (r-l)/2;
//treeIndex的節點分爲[l...mid]和[mid+1...r]兩部分
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
if (index>=mid/-1)
set(rightTreeIndex,mid+1,r,index,e);
else //index<= mid
set(leftTreeIndex,l,mid,index,e);
}
307號問題的解答
public class NumArray4 {
private SegmentTree<Integer> segTree;
public NumArray4(int[] nums) {
if(nums.length != 0){
Integer[] data = new Integer[nums.length];
for(int i = 0 ; i < nums.length ; i ++)
data[i] = nums[i];
segTree = new SegmentTree<>(data, (a, b) -> a + b);
}
}
public void update(int i, int val) {
if(segTree == null)
throw new IllegalArgumentException("Error");
segTree.set(i, val);
}
public int sumRange(int i, int j) {
if(segTree == null)
throw new IllegalArgumentException("Error");
return segTree.query(i, j);
}
}
9-7 更多線段樹相關的話題
線段樹的相關問題都比較困難,是高級數據結構。一般面試不會考,更多應用於算法競賽。
更多延伸
- 對於一個區間進行更新,例如將[2,5]去加中的所有元素+3
- 懶惰更新,使用lazy數組記錄未更新的內容。
- 二維線段樹
- 動態線段樹
- 區間操作相關另外一個重要數據結構:樹狀數組
- 區間相關問題:RMQ Range Minimum Query
最後
附上代碼完整定義
public interface Merger<E> {
//這裏用戶定義兩個區間如何融合的規則
E merge(E a,E b);
}
public class SegmentTree<E> {
private E[] tree;
private E[] data;
private Merger<E> merger;
public SegmentTree(E[] arr, Merger<E> merger){
this.merger = merger;
data = (E[]) new Object[arr.length];
for (int i=0;i<arr.length;i++)
data[i] = arr[i];
tree = (E[]) new Object[4 * arr.length];
buildSegmentTree(0,0,data.length-1);
}
//在treeIndex的位置創建表示區間[1....r]的線段樹
private void buildSegmentTree(int treeIndex,int l,int r){
if (l==r){
tree[treeIndex] = data[l];
return;
}
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
//int mid = (l+r)/2 爲避免l和r都特別大的時候使得l+r產生整形溢出的情況
int mid = l + (r-l)/2;
buildSegmentTree(leftTreeIndex,l,mid);
buildSegmentTree(rightTreeIndex,mid+1,r);
tree[treeIndex] = merger.merge(tree[leftTreeIndex],tree[rightTreeIndex]);
}
public int getSize(){
return data.length;
}
public E get(int index){
if (index<0 || index>=data.length)
throw new IllegalArgumentException("Index is illegal.");
return data[index];
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的左孩子節點的索引
private int leftChild(int index){
return 2*index + 1;
}
// 返回完全二叉樹的數組表示中,一個索引所表示的元素的右孩子節點的索引
private int rightChild(int index){
return 2*index + 2;
}
//返回區間[queryL,queryR]的值
public E query(int queryL, int queryR){
if (queryL<0 || queryL>=data.length || queryR<0 || queryR>=data.length || queryL>queryR)
throw new IllegalArgumentException("Index is illegal");
return query(0,0,data.length-1,queryL,queryR);
}
//在以treeIndex爲根的線段樹中[l,,,,r]的範圍裏,搜索區間[queryL....qeuryR]的值
private E query(int treeIndex,int l,int r,int queryL,int queryR){
if ( l==queryL && r == queryR)
return tree[treeIndex];
int mid = l+(r-l)/2;
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
if (queryL>=mid+1)
return query(rightTreeIndex,mid+1,r,queryL,queryR);
else if (queryR <= mid)
return query(leftTreeIndex,l,mid,queryL,queryR);
E leftResult = query(leftTreeIndex,l,mid,queryL,mid);
E rightResult = query(rightTreeIndex,mid+1,r,mid+1,queryR);
return merger.merge(leftResult,rightResult);
}
//將index位置的值,更新爲e
public void set(int index,E e){
if (index<0 || index >= data.length)
throw new IllegalArgumentException("Index is illegal");
data[index] = e;
set(0,0,data.length-1,index,e);
}
//在以treeIndex爲根的線段樹中更新index的值爲e
private void set(int treeIndex,int l,int r,int index,E e){
if (l==r){
tree[treeIndex] = e;
return;
}
int mid = l+ (r-l)/2;
//treeIndex的節點分爲[l...mid]和[mid+1...r]兩部分
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
if (index>=mid+1)
set(rightTreeIndex,mid+1,r,index,e);
else //index<= mid
set(leftTreeIndex,l,mid,index,e);
tree[treeIndex] = merger.merge(tree[leftTreeIndex],tree[rightTreeIndex]);
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
res.append('[');
for (int i=0;i<tree.length;i++){
if (tree[i]!=null)
res.append(tree[i]);
else
res.append("null");
if (i!=tree.length-1)
res.append(", ");
}
res.append(']');
return res.toString();
}
}