线段树(区间树)

为什么使用线段树？

什么是线段树？

每一个节点存储的是一个区间中相应统计值

最坏情况“n=2的k次方+1” 举例如下：可以看到叶子节点存在很多null节点，而也就是说存在一定的浪费，但为了满子元素全部存放只能这样。

查询操作

全局代码实现

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);
    }

    // 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);

        // int mid = (l + r) / 2;
        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 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 mid = 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 (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);
    }

    @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();
    }
}