数据结构-线段树

9、线段树

为什么使用线段树

最经典的线段染色问题

什么是线段树

线段树中每个节点存储的均是一个区间内的数字总和

• 叶子节点均在最后一层

• 叶子节点不一定都在最后一层

平衡二叉树

平衡二叉树中：二叉树中最大的深度和最小的深度之差最大等于1

线段树是平衡二叉树

堆也是平衡二叉树

线段树的实现

• 线段树的理论基础（将线段树看成是一个满二叉树，将有n个元素的数组构造成一个含有4n个节点的线段树）

import java.util.Arrays;

public class SegmentTree<E> {
    private E[] data;
    private E[] tree;
    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);
    }

    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];
    }

    private int leftChild(int index) {
        return 2*index+1;
    }

    private int rightChild(int index) {
        return 2*index+2;
    }

    private void buildSegmentTree(int treeIndex,int l ,int r) {
        if( l == r) {
            tree[treeIndex] = data[l];
            return;
        }

        int leftTreeSegment = leftChild(treeIndex);
        int rightTreeSegment = rightChild(treeIndex);
        int mid = l +(r-l)/2;
        buildSegmentTree(leftTreeSegment,l,mid);
        buildSegmentTree(rightTreeSegment,mid+1,r);
        tree[treeIndex] = merger.merge(tree[leftTreeSegment],tree[rightTreeSegment]);
    }

    @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();
    }
}

接口定义，用于动态设计左右子节点的具体任务，可以是左右节点之和，也可以是之差等等…

public interface Merger<E> {
    E merge(E l,E r);
}

测试程序

//测试线段树
Integer nums[] = {-2,-1,0,1,2,3};
//参数中添加匿名内部类Merger，或者直接通过lambda表达式:(a,b) -> a + b
SegmentTree<Integer> segmentTree = new SegmentTree<>(nums, new Merger<Integer>() {
    @Override
    public Integer merge(Integer l, Integer r) {
        return l +r;
    }
});
System.out.println(segmentTree);

查询线段树

//查询线段树某个区间内的值[queryL,queryR]
public E query(int queryL,int queryR) {
    if(queryL <0 || queryL >= data.length
        || queryR <0||queryR >= data.length) {
        throw new IllegalArgumentException("Index is illggal");
    }
     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 leftIndex = leftChild(treeIndex);
    int rightIndex = rightChild(treeIndex);
    if(queryL >= mid +1) {
        return query(rightIndex,mid+1,r,queryL,queryR);
    } else if(queryR <= mid) {
        return query(leftIndex,l,mid,queryL,queryR);
    }
    E leftReuslt = query(leftIndex,l,mid,queryL,mid);
    E rightReuslt = query(rightIndex, mid+1,r, mid+1,queryR);
    return merger.merge(leftReuslt,rightReuslt);
}

leetcode线段树相关操作

• 区域和检索-不可变

[0,1,2,3,4,5]------>[2,4] = 9

• 实现1

Class NumArray{
    private int[] sum;
    public NumArray(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];
    }
    
}

• 区域去检索-可变

Class NumArray{
    private int[] sum;
    private int[] data;
    public NumArray(int[] nums) {
        data = new int[nums.length];
        for(int i=0; i<nums.length;i++) {
            data[i] = nums[i];
        }
        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];
        }
    }
    
    //更新数据的操作过程
    public void update(int index,int val) {
        data[index] = val;
        for(int i =index+1; i<sum.length; i++) {
            sum[i] = sum[i-1]+data[i-1];
        }
    }
    
    public int sumRange(int i, int j) {
        return sum[j+1] -sum[i];
    }
    
}

线段树中的更新操作

//通过更新数组的值，更新线段树的节点
public void set(int index,E val) {
    if(index<0 || index>=data.length) {
        throw new IllegalArgumentException("Index is illegal");
    }
    data[index] = val;
}

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]);
}