Problem
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.
Example
For array [1,2,7,8,5], and queries [(0,4),(1,2),(2,4)], return [23,9,20]
Challenge
O(logN) time for each query
Note
很简单的题目,不过要达到O(logn)的时间,只能用前缀和的方法来做。
Solution
1. muggle法
public class Solution {
public ArrayList<Long> intervalSum(int[] A,
ArrayList<Interval> queries) {
ArrayList<Long> res = new ArrayList<Long> ();
for (int i = 0; i < queries.size(); i++) {
int start = queries.get(i).start;
int end = queries.get(i).end;
res.add(sum(A, start, end));
}
return res;
}
public long sum(int[] A, int start, int end) {
long sum = 0;
for (int i = start; i <= end; i++) {
sum += A[i];
}
return sum;
}
}
2. 前缀和法(推荐)
public class Solution {
public ArrayList<Long> intervalSum(int[] A,
ArrayList<Interval> queries) {
ArrayList<Long> res = new ArrayList<Long>();
long[] preSum = new long[A.length];
long sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
preSum[i] = sum;
}
for (Interval i: queries) {
if (i.start == 0) {
res.add(preSum[i.end]);
}
else {
res.add(preSum[i.end] - preSum[i.start-1]);
}
}
return res;
}
}
**粗体** _斜体_ [链接](http://example.com) `代码` - 列表 > 引用
。你还可以使用@
来通知其他用户。