*[Lintcode]Interval Sum区间最小数

Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers 

[start,
end]

. For each query, calculate the sum number between index start and end in the given array, return the result list.


 Notice

We suggest you finish problem Segment
Tree Build, Segment Tree Query andSegment
Tree Modify first.

Example

For array 

[1,2,7,8,5]

,
and queries 

[(0,4),(1,2),(2,4)]

, return 

[23,9,20]

使用区间树解决。注意interval各有一部分处于root左侧和右侧的问题。

/**
* Definition of Interval:
* public classs Interval {
* int start, end;
* Interval(int start, int end) {
* this.start = start;
* this.end = end;
* }
*/
public class Solution {
/**
*@param A, queries: Given an integer array and an query list
*@return: The result list
*/
public ArrayList<Long> intervalSum(int[] A,
ArrayList<Interval> queries) {
// write your code here
SegmentTreeNode node = build(A, 0, A.length - 1);
ArrayList<Long> res = new ArrayList<Long>();

for(Interval i : queries) {
res.add(find(node, i));
}
return res;
}

private Long find(SegmentTreeNode root, Interval i) {
if(root == null) return 0L;
if(i.start > root.end || i.end < root.start) return 0L;

if(i.start <= root.start && i.end >= root.end) {
return root.sum;
}
else {
//should caculate separately to deal with separate interval problem
Long left = find(root.left, i);
Long right = find(root.right, i);
return left + right;
}
}

private SegmentTreeNode build(int[] A, int start, int end) {
if(start > end) return null;

SegmentTreeNode root = new SegmentTreeNode(start, end);
if(start == end) {root.sum = (long)A[start]; return root;}

root.left = build(A, start, (start + end) / 2);
root.right = build(A, (start + end) / 2 + 1, end);

root.sum = root.left.sum + root.right.sum;
return root;
}

class SegmentTreeNode {
SegmentTreeNode left;
SegmentTreeNode right;
int start, end;
Long sum;
public SegmentTreeNode(int start, int end) {
this.start = start;
this.end = end;
}
}
}