Lintcode: Interval Sum

Posted 2020-06-13

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.

Have you met this question in a real interview? Yes
Example
For array [1,2,7,8,5], and queries [(0,4),(1,2),(2,4)], return [23,9,20]

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

Challenge
O(logN) time for each query

这道题最简便的方法当然是prefix Sum

/**
 * 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
        long[] prefixSum = new long[A.length+1];
        for (int i=0; i<A.length; i++) {
            prefixSum[i+1] = prefixSum[i] + A[i];
        }
        ArrayList<Long> res = new ArrayList<Long>();
        for (Interval interval : queries) {
            int start = interval.start;
            int end = interval.end;
            long result = prefixSum[end+1] - prefixSum[start];
            res.add(result);
        }
        return res;
    }
}

用Segment Tree

 1 /**
 2  * Definition of Interval:
 3  * public classs Interval {
 4  *     int start, end;
 5  *     Interval(int start, int end) {
 6  *         this.start = start;
 7  *         this.end = end;
 8  *     }
 9  */
10 public class Solution {
11     /**
12      *@param A, queries: Given an integer array and an query list
13      *@return: The result list
14      */
15     public class SegmentTreeNode {
16         long sum;
17         int start;
18         int end;
19         SegmentTreeNode left;
20         SegmentTreeNode right;
21         public SegmentTreeNode(int start, int end) {
22             this.sum = 0;
23             this.start = start;
24             this.end = end;
25             this.left = null;
26             this.right = null;
27         }
28     }
29      
30     SegmentTreeNode root;
31      
32     public ArrayList<Long> intervalSum(int[] A, 
33                                        ArrayList<Interval> queries) {
34         // write your code here
35         ArrayList<Long> res = new ArrayList<Long>();
36         root = build(A, 0, A.length-1);
37         for (Interval interval : queries) {
38             int start = interval.start;
39             int end = interval.end;
40             res.add(query(root, start, end));
41         }
42         return res;
43     }
44     
45     public SegmentTreeNode build(int[] A, int start, int end) {
46         SegmentTreeNode cur = new SegmentTreeNode(start, end);
47         if (start == end) {
48             cur.sum = A[start];
49         }
50         else {
51             int mid = (start + end)/2;
52             cur.left = build(A, start, mid);
53             cur.right = build(A, mid+1, end);
54             cur.sum = cur.left.sum + cur.right.sum;
55         }
56         return cur;
57     }
58     
59     public Long query(SegmentTreeNode cur, int start, int end) {
60         if (cur.start==start && cur.end==end) return cur.sum;
61         int mid = (cur.start + cur.end)/2;
62         if (end <= mid) return query(cur.left, start, end);
63         else if (start > mid) return query(cur.right, start, end);
64         else return query(cur.left, start, mid) + query(cur.right, mid+1, end);
65     }
66 }