难度:Easy
给定一个整数数组 nums,求出数组从索引 i 到 j (i ≤ j) 范围内元素的总和,包含 i, j 两点。
示例:给定 nums = [-2, 0, 3, -5, 2, -1],求和函数为 sumRange()sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3说明:你可以假设数组不可变。会多次调用 sumRange 方法。
简单题:
class NumArray {private:vector<int> num;public:NumArray(vector<int>& nums) {num=nums;}int sumRange(int i, int j) {return accumulate(num.begin()+i, num.begin()+j+1,0);}};/*** Your NumArray object will be instantiated and called as such:* NumArray* obj = new NumArray(nums);* int param_1 = obj->sumRange(i,j);*/