线段树

线段树不是一棵完全二叉树，因为区间可能不能完全平分，但线段树是一棵平衡二叉树

对一棵满二叉树，h层，一共有2^h-1个节点，对于h-1层，有2^(h-1)个节点，最后一层的节点数比前面所有层节点数之和多一

即如果区间有n个元素的话，最多需要4n的空间就可以全部放下（做了较大的富裕），且线段树不考虑添加元素，区间是固定的，我们使用4n的静态空间就可以了

SegmentTree

package com.liuyao;


/**
 * @author liuyao
 * @date 2018/07/28
 */
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, arr.length - 1);
    }

    /**
     * 构造线段树
     *
     * @param treeIndex
     * @param l
     * @param r
     */
    private void buildSegmentTree(int treeIndex, int l, int r) {
        // 递归截止条件，最后只左右相等，将当前的元素赋给tree[treeIndex]
        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.merge(tree[leftTreeIndex], tree[rightTreeIndex]);
    }

    /**
     * 获得数组中的元素个数
     *
     * @return
     */
    public int getSize() {
        return data.length;
    }

    /**
     * 根据索引返回相应的数组中的值
     *
     * @param index 索引
     * @return
     */
    public E get(int index) {
        if (index < 0 || index >= data.length) {
            throw new IllegalArgumentException("Index is illegal");
        }
        return data[index];
    }

    /**
     * 返回一棵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
     *
     * @param index
     * @return
     */
    public int leftChild(int index) {
        return 2 * index + 1;
    }

    /**
     * 返回一棵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
     *
     * @param index
     * @return
     */
    public int rightChild(int index) {
        return 2 * index + 2;
    }

    /**
     * 返回区间[queryL,queryR]的值
     *
     * @param queryL
     * @param queryR
     * @return
     */
    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[query...queryR]的值
     *
     * @param treeIndex
     * @param l
     * @param r
     * @param queryL
     * @param queryR
     * @return
     */
    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;
//        找到以treeIndex为根左右两个孩子的索引
        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);
    }

    /**
     *  将index位置的值，更新为e
     * @param index
     * @param 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
     * @param treeIndex
     * @param l
     * @param r
     * @param index
     * @param 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 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.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();
    }
}

Merger

package com.liuyao;

/**
 * @author liuyao
 * @date 2018/07/28
 */
public interface Merger<E> {
    public E merge(E a, E b);
}

Main

package com.liuyao;

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);

        System.out.println(segTree.query(2,5));
    }
}