玩转数据结构 （第八章 线段树）

线段树（区间树）
对于给定区间
更新：更新区间中一个元素或者一个区间的值
查询：查询一个区间[i,j]的最大值，最小值或者区间数字和
线段树是平衡二叉树（对于整棵树来说，最大深度和最小深度相差不超过一）（堆也是平衡二叉树）
“
如果区间有n个元素，数组表示需要有多少节点？
0层：1
1层：2
2层：4
3层：8
…
h-1层：2^(h-1)
”
对满二叉树：
h层，一共有2^{h-1个节点（大约是2}h）
最后一层（h-1层），有2^(h-1)个节点
最后一层的节点数大致等于前面所有层节点之和
需要大概4n的空间（可能会浪费一些空间）
代码实现
接口：

public interface Merger<E>{
	E merger(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的位置创建表示区间[l...r]
	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 - l) / 2;
		buildSegmentTree(leftTreeIndex, l, mid);
		buildSegmentTree(rightTreeIndex, mid + 1, r);
	
		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 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
	private int rightChild(int index){
		return 2 * index + 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 length.");

		return query(0, 0, data.length - 1, queryL, queryR);
	}

	//在以treeID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 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.merger(leftResult, rightResult);
	}

	@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(']');
			}
		}
		return res.toString();
	}

	//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);
	}

	// 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[index] = e;
			return;
		}

		int mid = l + (r - l) / 2;
		int leftTreeIndex = leftChild(treeIndex);
		int rightTreeIndex = rightChild(treeIndex);
		if (index >= mid + 1)
			set(rightTreeIndex, mid + 1, r, index, e);
		else
			set(leftTreeIndex, l, mid, index, e);

		tree[treeIndex] = merger.merger(tree[leftTreeIndex], tree[rightTreeIndex]);
	}

	@Override
	public String toString(){
		StringBuilder res = new StringBuilder();
		res.add('[');
		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(']');
			}
		}
	}
}

Main函数
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));
	}
}