Range Sum Query

Range Sum Query - Immutable

题目来源：https://leetcode.com/problemset/algorithms/
题目类型：动态规划

-题目描述-
-解题思路-
-代码实现-

题目描述

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

解题思路

题意为计算数组中下标大于等于i,小于等于j的元素的和。只需要将numt累加就可以得到结果。

代码实现

@requires_authorizationclass NumArray {public:    NumArray(vector<int> nums) {        for (int i = 0; i < nums.size(); i++) {            this->nums.push_back(nums[i]);        }    }    int sumRange(int i, int j) {        if (i > j) return 0;        if (i == j) return nums[i];        int sum = 0;        for (int t = i; t <= j; t++) {            sum += nums[t];        }        return sum;    }    private:    vector<int> nums;};