LeetCode:Range Sum Query - Immutable

walker lee

于 2016-05-11 20:40:29 发布

阅读量3.3k

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode OJ LeetCode解题记录文章标签： LeetCode Range Sum Query - Im

本文链接：https://blog.youkuaiyun.com/itismelzp/article/details/51378075

LeetCode OJ 同时被 2 个专栏收录

216 篇文章

订阅专栏

LeetCode解题记录

216 篇文章

订阅专栏

本文介绍了一个简单的区间求和查询问题及解决方案。对于给定的整数数组，文章提供了高效的算法来计算任意两个索引之间的元素总和。通过预处理数组元素的累积和，可以快速响应多次区间求和请求。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Range Sum Query - Immutable

Total Accepted: 29091 Total Submissions: 118678 Difficulty: Easy

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

You may assume that the array does not change.
There are many calls to sumRange function.

Subscribe to see which companies asked this question

Hide Tags

Dynamic Programming

Hide Similar Problems

(M) Range Sum Query 2D - Immutable (M) Range Sum Query - Mutable (E) Maximum Size Subarray Sum Equals k

c++ code:

class NumArray {
public:
    NumArray(vector<int> &nums) {
        sums.push_back(0);
        for(int i=0;i<nums.size();i++) {
            sums.push_back(sums.back() + nums[i]);
        }
    }

    int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
private:
    vector<int> sums;
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);