Medium Segment Tree Query
34%
Accepted
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
Have you met this question in a real interview? 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.
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
Note
It is much easier to understand this problem if you finished Segment Tree Build first.
/**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, max;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int max) {
* this.start = start;
* this.end = end;
* this.max = max
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param root, start, end: The root of segment tree and
* an segment / interval
*@return: The maximum number in the interval [start, end]
*/
public int query(SegmentTreeNode root, int start, int end) {
// write your code here
if(root == null || root.start > end || root.end < start) return 0;
if(root.start == root.end) return root.max;
return Math.max(query(root.left, start, end), query(root.right, start, end));
}
}
II===>
//**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, count;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int count) {
* this.start = start;
* this.end = end;
* this.count = count;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param root, start, end: The root of segment tree and
* an segment / interval
*@return: The count number in the interval [start, end]
*/
public int query(SegmentTreeNode root, int start, int end) {
// write your code here
if(root == null || root.start > end || root.end < start) return 0;
if(root.start == root.end) return root.count;
return query(root.left, start, end)+ query(root.right, start, end);
}
}
II===>
For an array, we can build a
SegmentTree
for it, each node stores an extra attribute count
to denote the number of elements in the the array which value is between interval start and end. (The array may not fully filled by elements)
Design a
Have you met this question in a real interview? query
method with three parameters root
, start
and end
, find the number of elements in the in array's interval [start, end] by the given root of value SegmentTree.
Yes
Example
For array
[0, 2, 3]
, the corresponding value Segment Tree is: [0, 3, count=3]
/ \
[0,1,count=1] [2,3,count=2]
/ \ / \
[0,0,count=1] [1,1,count=0] [2,2,count=1], [3,3,count=1]
query(1, 1)
, return 0
query(1, 2)
, return 1
query(2, 3)
, return 2
query(0, 2)
, return 2
Note
Tags Expand
It is much easier to understand this problem if you finished Segment Tree BuildandSegment Tree Query first.
//**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, count;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int count) {
* this.start = start;
* this.end = end;
* this.count = count;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param root, start, end: The root of segment tree and
* an segment / interval
*@return: The count number in the interval [start, end]
*/
public int query(SegmentTreeNode root, int start, int end) {
// write your code here
if(root == null || root.start > end || root.end < start) return 0;
if(root.start == root.end) return root.count;
return query(root.left, start, end)+ query(root.right, start, end);
}
}
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