【數據結構】—— 7、線段樹(區間樹)

爲什麼要使用線段樹?
實質 : 基於區間的統計查詢

一個節點存儲的是一個區間相應的和

public class SegmentTree<E> {

    private E[] tree;
    private E[] data;
    private Merger<E> merger;

    public SegmentTree(E[] arr) {
        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的位置創建表示區間[l...r]的線段樹
    private void buildSegmentTree(int treeIndex, int l, int r) {

        if (l == r) { // 此時只有一個元素
            tree[treeIndex] = data[l];
            return;
        }

        int leftTreeIndex = leftChild(treeIndex);
        int rightTreeIndex = rightChild(treeIndex);

        // int mid = (l + r) / 2;
        int mid = l + (r - l) / 2;
        // 兩個區間
        buildSegmentTree(leftTreeIndex, l, mid);
        buildSegmentTree(rightTreeIndex, mid + 1, r);

        // 具體操作合併的值和業務相關
        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];
    }

    // 返回完全二叉樹的數組表示中，一個索引所表示的元素的左孩子節點的索引
    private int leftChild(int index) {
        return 2 * index + 1;
    }

    // 返回完全二叉樹的數組表示中，一個索引所表示的元素的右孩子節點的索引
    private int rightChild(int index) {
        return 2 * index + 2;
    }


    // 返回區間[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爲根的線段樹中[l...r]的範圍裏，搜索區間[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的節點分爲[l...mid]和[mid+1...r]兩部分

        int leftTreeIndex = leftChild(treeIndex);
        int rightTreeIndex = rightChild(treeIndex);
        if(queryL >= mid + 1) // 只和右節點有關
            return query(rightTreeIndex, mid + 1, r, queryL, queryR);
        else if(queryR <= mid) // 只和左節點有關
            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);
    }
}