区域和检索 - 数组不可变(线段树求解)

给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。

示例：

给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
说明:

你可以假设数组不可变。
会多次调用 sumRange 方法。

代码如下:

public class NumArray {
	private SegmentTree<Integer> segmentTree;

	public NumArray(int[] nums) {
		if (nums.length > 0) {
			Integer[] data = new Integer[nums.length];
			for (int i = 0; i < nums.length; i++)
				data[i] = nums[i];
			segmentTree = new SegmentTree<>(data, (a, b) -> a + b);
		}
	}  

	public int sumRange(int i, int j) {
		if(segmentTree==null)
			throw new IllegalArgumentException("Segment Tree is null");
		return segmentTree.query(i, j);
	}
}

Merger接口：

public interface Merger<E> {
	E merge(E a,E b);
	
}

线段树:

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]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 - 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 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结点的索引
	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.merge(leftResult, rightResult);
	}

}