Segment Tree

ST基本表示

平衡二叉树定义(AVL)：它或者是一颗空树，或者具有以下性质的二叉树：它的左子树和右子树的深度之差(平衡因子)的绝对值不超过1，且它的左子树和右子树都是一颗平衡二叉树。

创建ST

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; 防止溢出
        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;
    }

    @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();
    }
}

查询

 // 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);
    }

303 Leetcode 区域和检索 不可变

使用线段树的解题思路

/// 303. Range Sum Query - Immutable
/// https://leetcode.com/problems/range-sum-query-immutable/description/

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);
    }
}

不使用线段树的解题思路（预处理）

/// 303. Range Sum Query - Immutable
/// https://leetcode.com/problems/range-sum-query-immutable/description/

public class NumArray2 {

    private int[] sum; // sum[i]存储前i个元素和, sum[0] = 0
                       // 即sum[i]存储nums[0...i-1]的和
                       // sum(i, j) = sum[j + 1] - sum[i]
    public NumArray2(int[] nums) {

        sum = new int[nums.length + 1];
        sum[0] = 0;
        for(int i = 1 ; i < sum.length ; i ++)
            sum[i] = sum[i - 1] + nums[i - 1];
    }

    public int sumRange(int i, int j) {
        return sum[j + 1] - sum[i];
    }
}

对于一个区间进行更新

因为遍历节点几乎是全部节点，所以约等于O（n）复杂度，所以可以采用懒惰更新 ，不去更新叶子节点，等下次使用的时候再更新

树状数组（Binary Index Tree）

RMQ(Range Minimum Query)