Lintcode: Segment Tree Query

Posted 2020-06-12

For an integer array (index from 0 to n-1, where n is the size of this array), in the corresponding SegmentTree, each node stores an extra attribute max to denote the maximum number in the interval of the array (index from start to end).

Design a query method with three parameters root, start and end, find the maximum number in the interval [start, end] by the given root of segment tree.

Have you met this question in a real interview? Yes
Example
For array [1, 4, 2, 3], the corresponding Segment Tree is:

                  [0, 3, max=4]
                 /                       [0,1,max=4]        [2,3,max=3]
          /         \        /            [0,0,max=1] [1,1,max=4] [2,2,max=2], [3,3,max=3]
query(root, 1, 1), return 4

query(root, 1, 2), return 4

query(root, 2, 3), return 3

query(root, 0, 2), return 4

 

这道题的启示是：Segment Tree 不光是可以找线段和，还可以找线段max, min等各种各样的指标

 1 /**
 2  * Definition of SegmentTreeNode:
 3  * public class SegmentTreeNode {
 4  *     public int start, end, max;
 5  *     public SegmentTreeNode left, right;
 6  *     public SegmentTreeNode(int start, int end, int max) {
 7  *         this.start = start;
 8  *         this.end = end;
 9  *         this.max = max
10  *         this.left = this.right = null;
11  *     }
12  * }
13  */
14 public class Solution {
15     /**
16      *@param root, start, end: The root of segment tree and 
17      *                         an segment / interval
18      *@return: The maximum number in the interval [start, end]
19      */
20     public int query(SegmentTreeNode root, int start, int end) {
21         // write your code here
22         if (root.start==start && root.end==end) return root.max;
23         int mid = (root.start+root.end)/2;
24         if (end <= mid) return query(root.left, start, end);
25         else if (start > mid) return query(root.right, start, end);
26         else return Math.max(query(root.left, start, mid), query(root.right, mid+1, end));
27     }
28 }