java——线段树 SegmentTree

应用:

  区间染色

  区间查询

线段树不是完全二叉树,线段树是平衡二叉树

使用数组来实现线段树:存储空间为4n

以下是使用数组实现的静态线段树:

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tree Index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);
        
        //根据业务组合线段树
        tree[treeIndex] = merger.merge(tree[leftTreeIndex], tree[rightTreeIndex]);
    }
    
    
    public E get(int index) {
        if(index < 0 || index >= data.length) {
            throw new IllegalArgumentException("Index is illegal.");
        }
        return data[index];
    }
    
    public int getSize() {
        return data.length;
    }
    
    private int leftChild(int index) {
        return 2*index + 1;
    }
    
    private int rightChild(int index) {
        return 2*index + 2;
    }
    
    public E query(int queryL, int queryR) {
        if(queryL < 0 || queryL >= data.length || queryR < queryL){
            throw new IllegalArgumentException("Index is illegal.");    
        }
        return query(0, 0, data.length -1, queryL, queryR);
    }
    //查询线段树
    //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);
        }else {
            //这种情况下产生了两段线段树,需要进行融合
            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
    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);
    }
    
    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();
    }
    
}

 对于一个区间的更新:

  懒惰更新:使用lazy数组记录未更新的内容,下一次访问时先访问lazy数组,若有内容,更新后再访问即可。

动态线段树:

  使用链表实现

  节省空间

  可以不均等划分区间,便于实际应用

原文地址:https://www.cnblogs.com/gaoquanquan/p/9881294.html