RMQ问题的线段树解法

RMQ（Range Minimum Query)问题是计算一个输入数列A[0...n-1]从位置i到位置j之间的最小值，即RMQ[i,j]=min{A[k], k=i,i+1...j}。RMQ的解法有很多，比如Sparse Table（ST）算法（注意这个ST缩写不是指Segement Tree哦）和转化为特殊的+1/-1 RMQ的算法。为了查询的方便，RMQ算法需要对数列A进行预处理(preprocessing)，如果用分别表示RMQ算法的预处理复杂度和查询复杂度的话，用线段树(segment tree)解决RMQ问题的复杂度为。

线段树解法首先构建线段树（一棵二叉树），根结点对应区间0...n-1，左右子树通过下标中间值分界并且没有重叠区间值，叶结点对应单值区间。如果设线段树根结点的高度为0，那么整棵树的高度不超过floor(logn)+1，按照完全二叉树的做法，将每个结点按level order存入一维数组M，数组M中每个元素值为该结点对应的区间[i...j]的最小值下标值k。虽然不能像完全二叉树那样保证元素之间没有空隙，但是可以利用下标值2k+1和2k+2快速访问某个结点k的左右孩子。查询时需要深度搜索树的结点，因此复杂度为O(logn)，比起Sparse Table等算法的查询复杂度为O(1)要差一些。

实现：

/**
 * 
 * Using Segment Tree to solve RMQ problem
 * time complexity: 
 * 
 * RMQ(A)-Range Minimum Query on array A: find the minimum value/index in the range A[i..j]
 *  
 * 
 * 
 *  
 * Copyright (c) 2011 ljs (http://blog.csdn.net/ljsspace/)
 * Licensed under GPL (http://www.opensource.org/licenses/gpl-license.php) 
 * 
 * @author ljs
 * 2011-08-02
 *
 */ 
public class RMQ_SegmentTree {	
	//return the heap-like array of the segment tree 
	//(Note: the segment tree is not a complete tree, but we
	//use an array to simulate a complete tree, so the array has some gaps)
	public int[] buildSegmentTree(int[] A){
		int n = A.length;
		int heapHeight = (int)(Math.log(n)/Math.log(2))+1;//h=floor(logN)+1; h starts with 0.
		int mSize = (1<<(heapHeight+1))-1; //total nodes = 2^(h+1)-1
		
		int[] M = new int[mSize];
		
		buildSegmentTree(M, 0, A, 0, n-1);
		return M;
	}
	
	private void buildSegmentTree(int[] M,int node,int[] A,int i,int j){		
		if(i==j){
			M[node] = i;
		}else{
			int leftnode=2*node+1;
			int rightnode=leftnode+1;
			
			buildSegmentTree(M,leftnode,A,i,(i+j)/2);
			buildSegmentTree(M,rightnode,A,(i+j)/2+1,j);
			
			if(A[M[leftnode]]<=A[M[rightnode]]){
				M[node] = M[leftnode];
			}else{
				M[node] = M[rightnode];
			}
		}
	}
	
	//x..y is the query interval of RMQ
	public int query(int[] M,int[] A,int x,int y){
		int n=A.length;
		return query(M,0,A,0,n-1,x,y);
	}
	
	//x..y is the query interval of RMQ
	//i..j is the current interval of a segment sub-tree's root
	private int query(int[] M,int node,int[] A,int i,int j,int x,int y){		
		//if the query interval doesn't intersect the current interval 
		//return -1
		if (x>j || y=j){
			return M[node];
		}
		
		//split query interval
		int leftnode=2*node+1;
		int rightnode=leftnode+1;
		
		int mid = (i+j)/2;
		if(x>mid){
			//right branch
			return query(M,rightnode,A,mid+1,j,x,y);
		}else if(y<=mid){
			//left branch
			return query(M,leftnode,A,i,mid,x,y);
		}else{
			//mixed
			int p1 = query(M,leftnode,A,i,mid,x,y);
			int p2 = query(M,rightnode,A,mid+1,j,x,y);
			
			if(p1==-1)return p2;
			if(p2==-1)return p1;
			
			if(A[p1]<=A[p2])
				return p1;
			else
				return p2;
		}
	}

	public static void main(String[] args) {
		RMQ_SegmentTree rmqSegTree = new RMQ_SegmentTree();
	
		int[] A = new int[]{0,1,2,3,7,1,9,2,8,6};			
		int[] M = rmqSegTree.buildSegmentTree(A);
		
			
		//print RMQ table RMQ[i,j], each cell has the form: value/index
		System.out.println("RMQ table:");
		for(int x=0;x