本文分为三部分:
1.文章介绍
2.实现代码及介绍
3.测试代码及测试结果
一.文章介绍
本文主要是使用了Java语言实现一个将数组构造为一个可以更新及查询的线段树结构, 线段树主要是用于解决线段和区间问题, 是一种高级的数据结构.
二.实现代码
自定义生成方式接口:
// 使线段树可以自定义生成方式
public interface Merge {
E merge(E a, E b);
}
线段树类代码:
public class SegmentTree {
private E[] tree;
private E[] data;
private Merge merge;
public SegmentTree(E[]arr, Merge merge){
this.merge = merge;
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);
}
// 构造线段树结构
private void buildSegmentTree(int treeIndex, int l, int r) {
if (l == r){
tree[treeIndex] = data[l];
return;
}
int leftChildIndex = leftChild(treeIndex);
int rightChildIndex = rightChild(treeIndex);
int mid = l + (r - l) / 2;
buildSegmentTree(leftChildIndex,l,mid);
buildSegmentTree(rightChildIndex,mid + 1,r);
tree[treeIndex] = merge.merge(tree[leftChildIndex],tree[rightChildIndex]);
}
public int getSize(){
return data.length;
}
public E get(int index){
if (index < 0 && index > data.length-1)
throw new IllegalArgumentException("Index is Illegal");
return data[index];
}
// 返回完全二叉树的数组表示中, 一个索引所表示的元素的左孩子节点的索引
public int leftChild(int index){
return 2*index + 1;
}
// 返回完全二叉树的数组表示中, 一个索引所表示的元素的右孩子节点的索引
public int rightChild(int index){
return 2*index + 2;
}
// 在线段树中搜索区间[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("Error");
return query(0, 0, data.length-1, queryL ,queryR);
}
// 在以treeIndex为根的线段树中[l...r]的范围里, 搜索区间[queryL...queryR]的值
private E query(int treeIndex, int l, int r, int queryL, int queryR) {
if (queryL == l && queryR == r)
return tree[treeIndex];
int mid = l + (r - l) / 2;
int leftChildIndex = leftChild(treeIndex);
int rightChildIndex = rightChild(treeIndex);
if (queryL >= mid + 1)
return query(rightChildIndex,mid + 1, r, queryL
, queryR);
else if (queryR <= mid)
return query(leftChildIndex,l, mid, queryL
, queryR);
E leftRes = query(leftChildIndex, l, mid, queryL, mid);
E rightRes = query(rightChildIndex, mid + 1, r, mid + 1, queryR);
return merge.merge(leftRes, rightRes);
}
// 将index位置的值, 更新为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);
}
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;
int leftChildIndex = leftChild(treeIndex);
int rightChildIndex = rightChild(treeIndex);
if (index >= mid + 1)
set(rightChildIndex, mid + 1, r, index, e);
else
set(leftChildIndex, l,mid,index,e);
tree[treeIndex] = merge.merge(tree[leftChildIndex], tree[rightChildIndex]);
}
@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(',');
else
res.append(']');
}
return res.toString();
}
}
三.测试代码
public static void main(String[] args) {
Integer[] nums = {-2, 0, 3, -5, 2, -1};
SegmentTree segmentTree = new SegmentTree<>(nums, (a, b) -> a + b);
System.out.println(segmentTree.query(2, 5));
segmentTree.set(2, 10);
System.out.println(segmentTree.query(2, 5));
System.out.println(segmentTree.toString());
}