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<f(n),g(n)>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<O(n),O(logn)>。
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: <O(n),O(logn)> * * 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<i) return -1; //if the query interval entirely covers the current interval //return the current node's minimum index if(x<=i && 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<A.length;x++){ for(int y=0;y<x;y++){ System.out.format(" "); } for(int y=x;y<A.length;y++){ int p = rmqSegTree.query(M,A,x,y); System.out.format(" %d/%d",A[p],p); } System.out.println(); } } }