主要用来出来区间计算问题
public interface Merger<E> {
E merger(E a, E b);
}
public class SegmentTree<E> {
private E[] tree;
private E[] data;
private Merger<E> merger;
//创建数组的大小 如果数组的长度正好为 2^n 此时 tree的长度2n就可以了 但是在数组的长度大于2^n的情况下 tree的长度就应该为4n了
public SegmentTree(E[] arr, Merger<E> merger) {
this.merger = merger;
data = (E[]) arr;
tree = (E[]) new Object[4 * arr.length];
buildSegmentTree(0, 0, data.length - 1);
}
// 在treeIndex的位置创建表示区间[l...r]的线段树
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);
//计算中间节点
int middle = l + (r - l) / 2;
buildSegmentTree(leftTreeIndex, l, middle);
buildSegmentTree(rightTreeIndex, middle + 1, r);
// 通过子节点计算 当前节点
tree[treeIndex] = merger.merger(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
private int leftChild(int index) {
return index * 2 + 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
private int rightChild(int index) {
return index * 2 + 2;
}
// 返回区间[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("Index is illegal.");
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 (l == queryL && r == queryR) {
return tree[treeIndex];
}
int middle = l + (r - l) / 2;
// treeIndex的节点分为[l...mid]和[mid+1...r]两部分
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
//查询的结果全部在 左边 或者 右边 的情况
if (queryL >= middle + 1) {
return query(rightTreeIndex, middle + 1, r, queryL, queryR);
} else if (queryR <= middle) {
return query(leftTreeIndex, l, middle, queryL, queryR);
}
//查询的结果在 左边和右边 都有的情况
E lr = query(leftTreeIndex, l, middle, queryL, middle);
E rr = query(rightTreeIndex, middle + 1, r, middle + 1, queryR);
return merger.merger(lr, rr);
}
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为根的线段树中更新index的值为e
private void set(int treeIndex, int l, int r, int index, E e) {
if (l == r) {
tree[treeIndex] = e;
return;
}
int middle = l + (r - l) / 2;
// treeIndex的节点分为[l...mid]和[mid+1...r]两部分
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
if (index >= middle + 1) {
set(rightTreeIndex, middle + 1, r, index, e);
} else {
set(leftTreeIndex, l, middle, index, e);
}
//内容变了 重新赋值
tree[treeIndex] = merger.merger(tree[leftTreeIndex], tree[rightTreeIndex]);
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder();
sb.append("[");
for (int i = 0; i < tree.length; i++) {
if (tree[i] != null) {
sb.append(tree[i]);
} else {
sb.append("null");
}
if (i != tree.length - 1) {
sb.append(",");
}
}
sb.append("]");
return sb.toString();
}
}
测试:
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);
System.out.println(segTree.query(0,2));
System.out.println(segTree.query(2,5));
System.out.println(segTree.query(0,5));
}
}
输出
[-3,1,-4,-2,3,-3,-1,-2,0,null,null,-5,2,null,null,null,null,null,null,null,null,null,null,null]
1
-1
-3